Basic Calculus Unveiled1 - Sound ... - Sound · PDF file2 Lecture 18. INTEGRAL CALCULUS:...

0

Larissa Fradkin

Basic Calculus Unveiled …………………………………………………… 14 September 2013

1

Table of Contents I. INTRODUCTION .............................................................................................. 4

II. CONCEPT MAPS ............................................................................................. 5

III. LECTURES ....................................................................................................... 7

Lecture 13. CALCULUS: Sequences, Limits and Series........................................... 7

13.1 Sequences ......................................................................................................... 7

13.2 Limit of a sequence .......................................................................................... 10

13.3 Series ............................................................................................................... 14

13.4 Instructions for self-study ................................................................................. 15

Lecture 14. DIFFERENTIAL CALCULUS: Limits, Continuity and Differentiation of Real Functions of One Real Variable ........................................................................ 17

14.1 Limits ............................................................................................................... 17

14.2 Continuity of a function .................................................................................... 19

14.3 Differentiation of a real function of one real variable ........................................ 20

14.4 A historical note ............................................................................................... 22


Lecture 15. DIFFERENTIAL CALCULUS: Differentiation (ctd.) .............................. 24

15.1 Differentiation Table ......................................................................................... 24

15.2 Differentiation Rules ......................................................................................... 25

15.3 Decision Tree for Differentiation ...................................................................... 29

15.4 The higher order derivatives ............................................................................ 29

15.5 The partial derivatives ...................................................................................... 30

15.6 Applications of differentiation ........................................................................... 30


Lecture 16. DIFFERENTIAL CALCUUS: Sketching Graphs Using Analysis ........... 32

16.1 Stationary points .............................................................................................. 32

16.2 Increasing and decreasing functions................................................................ 32

16.3 Maxima and minima ......................................................................................... 33

16.4 Convex and concave functions ........................................................................ 37

16.5 Inflexion points ................................................................................................. 37

16.6 Sketching rational functions using analysis ...................................................... 38

16.7 Applications of rational functions and their graphs ........................................... 41


Lecture 17. Application of DIFFERENTIAL CALCULUS to Approximation of Functions: the Taylor and Maclaurin Series ............................................................. 43

17.1 Approximating a real function of a real variable using its first derivative .......... 43

17.2 The Maclaurin polynomials .............................................................................. 43

17.3 The Taylor polynomials .................................................................................... 44

17.4 The Taylor series ............................................................................................. 44

17.5 The Maclaurin series ........................................................................................ 45


2

Lecture 18. INTEGRAL CALCULUS: Integration of Real Functions of One Real Variable (Definite Integrals) ...................................................................................... 50

18.1 A definite integral ............................................................................................ 50

18.2 Notation ........................................................................................................... 51

18.3 Comparison of a definite Integral and derivative .............................................. 51

18.4 Examples of integrable functions ..................................................................... 51

18.5 The Mean Value Theorem ............................................................................... 52

18.6 A definite integral with a variable upper limit - function Φ(x) ............................ 53

18.7 The Fundamental Theorem of Calculus ........................................................... 54

18.8 Applications of integration ................................................................................ 54

18.9 A historical note ............................................................................................... 54

18.10 Instructions for self-study ............................................................................... 55

Lecture 19. INTEGRAL CALCULUS: Integration of Real Functions of One Real Variable (Indefinite Integrals) .................................................................................... 56

19.1 The relationship between differentiation and integration .................................. 56

19.2 An indefinite integral ........................................................................................ 57

19.3 Finding an indefinite integral ............................................................................ 58


Lecture 20. INTEGRAL CALCULUS: Advanced Integration Methods .................... 65

20.1 Integration of products of trigonometric functions ............................................ 65

20.2 Integration by parts (integration of products of different types of functions) .... 66

20.3 Partial fractions ................................................................................................ 68

20.4 Integration of rational functions ........................................................................ 69

20.5 Decision Tree for Integration ............................................................................ 70


Lecture 21. INTEGRAL CALCULUS: Applications of Integration ............................ 72

21.1 Mean value of a function .................................................................................. 72

21.2 Electrical and mechanicsl systems ................................................................. 73

21.3 Rotational systems ........................................................................................... 74

21.4 Instructions for self-study ................................................................................ 76

Lecture 22. Ordinary Differential Equations ............................................................ 77

22.1 Basic concepts ................................................................................................. 77

22.2 Order of a differential equation ........................................................................ 78

22.3 Linearity or non-linearity of a differential equation ........................................... 78

22.4 Differential equations with constant coefficients .............................................. 78

22.5 Homogeneous and inhomogeneous ODEs ...................................................... 78

22.6 The first order linear homogeneous equation with constant coefficients .......... 78

22.7 The initial value problems ................................................................................ 79

22.8 Balance equations in chemical engineering ..................................................... 79

22.9 Ordinary differential equations with complex coefficients ................................. 80

IV. SUMMARIES .................................................................................................... 83

Algebra Summary ..................................................................................................... 83

Functions Summary .................................................................................................. 88

Order of Operations Summary .................................................................................. 89

Quadratics Summary ................................................................................................ 90

Trigonometry Summary ............................................................................................ 91

3

Complex Numbers .................................................................................................... 93

Decision Tree For Solving Simple Equations ............................................................ 94

Sketching Graphs by Simple Transformations .......................................................... 95

Finding a Limit of a Sequence .................................................................................. 96

Differentiation Summary ........................................................................................... 97

Integration Summary................................................................................................. 97

V. GLOSSARY ...................................................................................................... 98

VI. STUDY SKILLS FOR MATHS ....................................................................... 104

VII. TEACHING METHODOLOGY (FAQs) ........................................................... 104

4

I. INTRODUCTION These notes are based on the lectures delivered by the author to engineering students of London South Bank University over the period of 16 years. This is a University of widening participation, with students coming from many different countries, many of them not native English speakers. Most students have limited mathematical background and limited time both to revise the basics and to study new material. A system has been developed to assure efficient learning even under these challenging restrictions. The emphasis is on systematic presentation and explanation of basic abstract concepts. The technical jargon is reduced to the bare minimum. Nothing gives a teacher a greater satisfaction than seeing a spark of understanding in the students’ eyes and genuine pride and pleasure that follows such understanding. The author’s belief that most people are capable of succeeding in - and therefore enjoying - the kind of mathematics that is taught at Universities has been confirmed many times by these subjective signs of success as well as genuine improvement in students’ exam pass rates. Interestingly, no correlation had ever been found at the Department where the author worked between the students’ qualification on entry and graduation. The book owns a lot to the authors’ students – too numerous to be named here - who talked to her at length about their difficulties and successes, e.g. see Appendix VII on Teaching Methodology. One former student has to be mentioned though – Richard Lunt – who put a lot of effort into making this book much more attractive than it would have been otherwise. The author can be contacted through her website www.soundmathematics.com. All comments are welcome and teachers can obtain there the copy of notes with answers to questions suggested in the text as well as detailed Solutions to suggested Exercises. The teachers can then discuss those with students at the time of their convenience. Good luck everyone!

5

II. CONCEPT MAPS Throughout when we first introduce a new concept (a technical term or phrase) or make a conceptual point we use the bold red font. We use the bold blue to verbalise or emphasise an important idea. Two major topics are covered in this course, Elementary Algebra and Basic Calculus. Two major topics are covered in these Notes, Differential Calculus and Integral Calculus. You can understand these topics best if you first study the Notes on Elementary Algebra and Functions. Here is a concept map of Elementary Algebra.

Main concepts Visualisation

Examples Full list

Applications

… integer real complex

Elementary ALGEBRA

variables

operations on variables

addition subtraction multiplication division raising to power extracting roots taking logs

number line complex plane (the Argand diagram)

solving algebraic equations (transposing the formulae) advanced mathematics, science, engineering, business, finance

Venn diagram

6

Here is a concept map of Basic Calculus. It is best to study it before studying any of the Calculus Lectures to understand where they are on the map. The better you see the big picture the faster you learn!

using tables using simple transforma-tions using pointwise operations using analysis

operations on functions

trig algebraic

diagram functions (real functions of one real variable)

elementary advanced

( )n Pn( )

e( )

ln ( )

sin( ) cos( ) tan( )

using graphs approximate calculations finding means, areas, volumes… solving differential equations solving integral equations advanced mathematics, science, engineering business, finance

+-, */ composition/ decomposition taking inverse

taking a limit differentiation integration

graphs

BASIC CALCULUS

Main concepts Visualisation Applications l

sketching methods

Two types Some types

Examples Full list

7

III. LECTURES Lecture 13. CALCULUS: Sequences, Limits and Series In Calculus we study functions and operations on functions (see Calculus Concept Map). Elementary operations on functions (algebraic operations and composition/decomposition) were covered in Lectures 4 - 7. We now introduce the first advanced operation on functions – taking a limit. We introduce it first for discrete functions called sequences. 13.1 Sequences A sequence is an ordered set of numbers xn, a discrete function x(n), where the argument is an integer n. This means that a discrete function is a function whose domain is the set of all integers or else its subset. Such argument is called the counter. There are different ways of writing up a sequence: {xn} or x1 , x2 , x3 , ...., xN (each symbol n, 1,

2, 3, …, N here can also be called an index or subscript). A sequence can be finite (having a fixed number of elements) or infinite (whatever index you specify there is always an element in the sequence with a greater index). Traditionally sequence elements are called terms but in these notes we reserve the word “term” for addends. 13.1.1. Defining a Sequence as Function of Counter

A sequence can be defined (described) by specifying a relationship between each element and its counter.

Examples: 1. Describe sequence 1, 2, 3, 4, ... using the above notation. Solution x1 = 1, x2 = 2, x3 = 3, …

Question: What is the relationship between the counter and the sequence element? Answer:

⇒ xn = n.

2. Describe sequence 1, 4, 9, 16, ... using the above notation. Solution x1 = 1, x2 = 4, x3 = 9, ...

Question: What is the relationship between the counter and the sequence element? Answer:

⇒ xn = n2.

13.1.2. Visualising a Sequence as a Function of Counter

If a sequence is defined (described) by specifying a relationship between each element and its counter it can be visualised as a sequence of points on the number line (see figure 13.1) or else as a discrete graph (see figure 13.2).

8

Figure 13.1. An example of a sequence xn, n = 0, 1, …, 5 visualised using the number line. The arrowed arcs are used to indicate the order of elements in the sequence.

Figure 13.2. An example of a graph of sequence xn, n = 0, 1, …, 10. http://upload.wikimedia.org/wikipedia/commons/8/88/Sampled.signal.svg 13.1.3 Applications of Sequences

A digital signal x (see figure 13.3 below) has the following characteristics:

1) it holds a fixed value for a specific length of time 2) it has sharp, abrupt changes

3) a preset number of values is allowed. http://www.privateline.com/manual/threeA.html

Figure 13.3. A typical digital signal. http://www.privateline.com/manual/threeA.htm Therefore, digital signals are examples of sequences.

t

x

0 1 2 3 4 5 6 7 8 9 10 n

x 15

10

5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x

9

13.1.4. Defining a Sequence via a Recurrence Relationship

A sequence element may be defined in another way, via a recurrence relation

),,...,,( 111 xxxfx nnn −+ =

where f may be a function of several arguments and not just one argument.

Examples:

1. Given a sequence xn = n2 we can change the functional description to a

recurrence relation xn + 1 = (n + 1)

2 = n

2 + 2n + 1

⇒ xn + 1 = xn + 2 nx + 1

When given such a recurrence the first element needs to be specified for us to be able to start evaluating other elements. 2. A Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, ... can be described via a recurrence relation

xn + 1 = xn + xn - 1.

When given such a recurrence the first two elements have to be specified for us to be able to evaluate other elements. Let us check that this is a correct description:

Question: What are x1 and x2?

Answer: Question: Does x3 satisfy the given recurrence relation and why? Answer: Question: Does x4 satisfy the given recurrence relation and why? Answer:

Question: Does x5 satisfy the given recurrence relation and why? Answer:

Thus, there are two ways to describe a sequence, 1. using a functional relation x(n), which specifies how each sequence element is

defined by its counter;

2. using a recurrence relation ),,...,,( 111 xxxfx nnn −+ = which specifies how each

sequence element is defined by previous sequence element(s).

10

Fibonacci sequences in nature When superimposed over the image of a nautilus shell we can see a Fibonacci sequence in nature:

http://munmathinnature.blogspot.com/2007/03/fibonacci-numbers.html Each of the small spirals of broccoli below follows the Fibonacci’s sequence.

http://www.pdphoto.org/PictureDetail.php?mat=pdef&pg=8232 13.2 Limit of a sequence Taking a limit of a sequence as n grows larger and larger without bounds is the first advanced operation on functions that we cover. In mathematics, instead of the phrase n grows larger and larger without bounds we use the shorthand ∞→n (verbalised as n tends to infinity). If it exists the outcome of applying this operation to a (discrete) function xn is

called lim xn and is either a number or else ±∞ (either +infinity or -infinity). Sometimes

instead of lim xn we write nn

x∞→

lim . However, usually, description ∞→n is understood and

not mentioned.

Note: ∞ is not a number but a symbol of a specific sequence behaviour, ∞+ means that the sequence increases without bounds and ∞− means that the sequence decreases without bounds - see the right column in the Table in Section 13.2.1 below.

11

13.2.1 Definition of a Limit of a Sequence

Examples: Sketch the sequence to find its limit:

1. xn = n

1→ 0 Verbalise: as n increases, xn tend (come closer and closer) to 0.

2. xn = −n

1→ 0 Verbalise: as n increases, xn tend (come closer and closer) to 0.

3. xn = 0)1(

1

→−

+n

n Verbalise: as n increases, xn tend (come closer and closer) to 0.

4. xn = n → ∞ Verbalise: as n increases, xn tend to ∞ (increase without bounds).

5. xn = − n → − ∞ Verbalise: as n increases, xn tend to - ∞ (decrease without bounds). Sometimes, limits can be found using analytical considerations or else using the Table,

Examples of a finite limit axn

=∞→

n)(

lim

(can also write

)( ∞→

→nn ax )

Examples of an infinite limit ±∞=∞→

n)(

lim xn

(can also write

)( ∞→

±∞→n

nx )

Example 1: −→ axn

or xn gets closer and closer to a from below, maybe

reaches it.

Example 1: →nx + ∞

or

For any number M there exists N : Mxn > for

all Nn ≥ (sequence increases without bounds)

Example 2: xn → a+ or xn gets closer and closer to a from above, maybe reaches it.

Example 2: →nx - ∞

or

For any number M there exists N : Mxn < for

all Nn ≥ (sequence decreases without bounds)

Example 3: xn → a x or xn gets closer and closer to a, maybe reaches it.

a x

x3

x2

x1 1 2 3 n

x1 x2 x3 a x 1 2 3 4 n

x4 x

x3

x2

x1

x1 x2 x3 x4 x

x x1

x2

x3

a 1 2 3 n a x3 x2 x1 x 1 2 3 4 n

x

x1

x2

x3

x4 x4 x3 x2 x1 x

x1 x3 x5 a x4 x2 x

x2

x4

a

x3

x1

a 1 2 3 4 n

12

Rules and a Decision Tree for Limits: 13.2.2 Table of Limits

If {xn} has a finite limit the sequence is said to converge. 13.2.3 Rules for (Properties of) Limits

1. The limit is unique. 2. lim (αxn + βyn) = α lim xn + β lim yn, α, β – constant Verbalise: limit of a sum is a sum of limits and any constant factor can be taken out – linearity property. 3. lim xn yn = lim xn lim yn Verbalise: limit of a product is a product of limits.

4. ,lim

limlim

n

n

n

n

y

x

y

x= lim yn ≠0

Verbalise: limit of a ratio is a ratio of limits. Laws 2 - 4 give a general recipe for evaluating a limit of a composition of sequences – instead of the sequence element substitute its limit. That is, the Laws 2 – 4 state that if a limit exists the operation of taking the limit and any subsequent algebraic operation can be performed in reverse order. For more complicated operations such reversal is not always justified.

Optional

5. xn ≥ yn ⇒ lim xn ≥ lim yn - every element of one sequence is greater than or equal to the element with the same index in the other sequence ⇒ its limit is also greater than or equal to the limit of the other sequence.

6. cxn ≤ yn ≤ zn ⇒ lim xn = lim zn ⇒ lim yn = lim xn a sequence squeezed between two others with the same limit has the same limit.

Example: .)( 1nx

nn

+−= lim xn ?

Solution: lim xn does not exist ⇐ Rule 1

xn lim xn

1/n 0

0a ,1

>na

1

qn , |q| < 1 0

n

n)

11( + e

easy to find

the proof is rather involved (e ≈ 2.71, Neper’s constant, an irrational number).

13

13.2.4 Indeterminate forms

Sometimes using the above rules leads to an indeterminate form (no obvious answer) like

0

0,

∞

∞, ∞⋅0 , ∞−∞ , 00, 1 ∞ etc. - MEMORISE

Question: Why do these outcomes present no clear answer?

Answer: 0

0

is indeterminate, because on the one hand, 0 divided by any number is 0 and

on the other hand, division of any number by smaller and smaller numbers leads to larger

and larger answers – tends to ∞ . ∞

∞ is indeterminate, because on the one hand, larger and

larger numbers divided by a number give a larger and larger result (the result tends to ∞ ) and on the other hand, any number divided by larger and larger numbers gives a smaller and smaller result (the result tends to 0) etc. When substitution leads to an indeterminate form we use tricks in an attempt to evaluate the limit or determine it to be non-existent.

Examples:

1. 2

1

nxn = ,

n

nn

y

x

ny lim

1⇒= =

n

n1

1

lim2

= 01

lim =n

2. 2

1

nxn = ,

n

nn

x

y

ny lim

1⇒= =

2

1

1

lim

n

n = ∞=nlim

3. nxn = , 2nyn =

)1(lim)lim()lim( 2 nnnnyx nn −=−=−⇒ = −∞=−∞⋅∞=− )()1lim(lim nn

4. 2nxn = , 12 −= nyn

2

2 1limlim

n

n

x

y

n

n −=⇒ =

1

11

lim1

lim

22

2

2 nn

n

n−

=− ÷

= 1 – 0 = 1

13.2.5 Decision Tree for Limits

Decision Tree (a Flow Chart) for evaluating limits is given in figure 13.1. Start reading it from the very top and follow the arrows as dictated by your answers. The tricks required to find a limit can be very complicated. However, in standard undergraduate courses they involve only simple algebraic and trigonometric laws and formulae.

0

0 use algebraic trick – FLIP RULE

0

0 use algebraic trick – FLIP RULE

∞

∞ use algebraic trick – divide top and bottom by the highest power

∞−∞ use algebraic trick - factorisation

14

Figure 13.1. Decision Tree for Limits. 13.3 Series A series is a sum of sequence elements. Question: What is a sum? Answer:

A series Nxx ++ ...1 can be written using the shorthand SN or ∑

=

N

n

nx

1

. In the latter case we

say that we use the sigma notation, because ∑ is a capital Greek letter called sigma.

13.3.1 Arithmetic Progression

An arithmetic progression is a sequence defined by the following recurrence relationship:

axx nn +=+1 , where a is a constant (with respect to n). a is called the common difference.

Answer

No Yes

Yes No

Maybe, tricks

Is one of the factors const 1≠ ?

What is the last operation in xk?

Is xk a table sequence?

Yes No

+ -

Other

Separate terms

constant factor

out

An indeterminate form?

(∞ − ∞, 0⋅∞, 0/0, ∞/∞, etc.)

If allowed reverse order of lim and subsequent operation

Use a trick

xk ?

x

÷

15

2... 1

1

N

NN

xxNxxS

+=++=⇒ MEMORISE

13.3.2 Geometric Progression

A geometric progression is a sequence defined by the following recurrence relationship:

nn axx =+1 , where where a – constant (with respect to n). a is called the common ratio.

a

axS

N

N−

−=⇒

1

)1(1

If │a│ < 1, then a

x

a

ax

a

xS

N

N−

→−

−−

=111

111 a

xSN

−=⇒

1lim 1 . MEMORISE

13.4 Instructions for self-study

• Revise Summaries on ALGEBRA, FUNCTIONS and TRIGONOMETRY

• Revise Lecture 11 and study Solutions to Exercises in Lecture 11 using the STUDY SKILLS Appendix

• Revise Lecture 12 using the STUDY SKILLS Appendix

• Study Lecture 13 using the STUDY SKILLS Appendix

• Attempt the following exercises: Q1. Find the following limits:

a) 12

1lim

2

+

++

n

nn

b) 12

1lim

2

2

+

++

n

nn

c) 12

1lim

3

2

+

++

n

nn

Q2. Find the sum of a hundred terms of the arithmetic series with first term 1 and common difference 2. Q3. A steel ball-bearing drops on to a smooth hard surface from a height h. The time to first

impact is ,2

g

hT = where g is the acceleration due to gravity. The times between

successive bounces are ,...,2 ,2 ,2 32 TeTeeT where e is the coefficient of restitution between

the ball and the surface ( 10 << e ). Find the total time taken up to the fifth bounce if T = 1 and e = 0.1. Q4 (advanced). The irrational number π can be defined as the area of a unit circle (circle of radius 1). Thus, it can be defined as a limit of a sequence of known numbers an. The method used by Archimedes was to inscribe in the unit circle a sequence of regular polygons (that is, polygons with equal sides – see figure 8.1). As the number of sides increases the polygon “fills” the circle. For any given number n find the area an of the regular polygon inscribed in the unit circle. Show, by use of trigonometric identities

16

θθθ cossin22sin = , θθ 2sin212cos −= and 1cossin 22 =+ θθ (Pythagoras’ theorem), that the area an satisfies the recurrence relation

Use trigonometry to show that a4 = 2 and use the above

recurrence relation to find a8.

Figure 8.1. A circle with two inscribed regular polygons, one with four sides (a square) and

one, with six (a hexagon).

Q5. Explain why ∑∑∑∞

=

∞

=

∞

=

+==011

]1[][][

knk

kxnxkx .

.4 ,2

112

222 ≥

−−=

n

n

a

n

a nn

17

Lecture 14. DIFFERENTIAL CALCULUS: Limits, Continuity and Differentiation of Real Functions of One Real Variable

14.1 Limits Taking a limit is an advanced operation on functions. The result is a number (more generally, a constant with respect to the independent variable). The operation involves finding a limiting value of the dependent variable when the independent variable approaches (tends to) a specified limiting value of its own or else to ∞ or ∞ . 14.1.1 Definition of a Limit of a Real Function of One Real Variable

There are four possible situations when evaluating the limit of a function:

1. cxfax

=→

)(lim

Verbalise: as x tends to a finite limit a, f(x) tends to a finite limit c.

2. cxfcxfxx

==−∞→∞→

)(limor )(lim

Verbalise: as x tends (goes off) to infinity, f(x) tends to a finite limit c. 6. Verbalise: as x tends (goes off) to + or - infinity, f(x) tends to + or - infinity.

4. =→

)(lim xfax

∞ or ∞-

Verbalise: as x tends to (approaches) a finite limit a, f(x) tends to + or - infinity.

Note: symbols ±±±±∞∞∞∞ are not numbers, they describe a specific function behaviour. Examples: 1.

0lim =−∞→

x

xe ∞=

∞→

x

xelim

,-or )(lim ,-or )(lim ∞∞=∞∞=−∞→∞→

xfxfxx

1

x

y = ex

a x

y = f(x)

c

18

⇒

2.

−∞=+→

xx

lnlim0

∞=∞→

xx

lnlim

14.1.2 Table of Limits of Functions

f(x) )(lim0

xfx→

x

xsin

1

x

ex

1− 1

14.1.3 Rules for (Properties of) Limits

1. The limit is unique (that is, there can be only one limiting value).

2. ax→

lim [f1 f2 (x)] = ax

lim→

f1 (x) ax

lim→

f2 (x) if all limits exist

Using rule 2 is equivalent to just substituting a for x. If substitution produces an indeterminate form tricks have to be used to establish whether the limit exists and if yes, what is it.

Optional

3. f1 (x) ≤ f2 (x) ≤ f3 (x) and ax

lim→

f1 (x) = ax

lim→

f3 (x)

axlim

→f2 (x) =

axlim

→f1 (x)

Examples:

1. 1x

lim→

(x2 – 4x + 8) = 1

2 – 4 · 1 + 8 = 5 (instead of x we substituted its limiting value)

2. 2

limπ→x

sin x = 1 (instead of x we substituted its limiting value)

3. 2

limπ→x

cos x = 0 (instead of x we substituted its limiting value)

+ -

× ÷

+ -

× ÷

the proof is rather involved

x

y = ln x

1

19

4. 2

limπ→x

tan x does not exist, but ±→

2

limπx

∞=mx

x

cos

sin, which is a short-hand for two

statements,+→

2

limπx

∞−=x

x

cos

sin and

−→2

limπx

∞+=x

x

cos

sin

5. 2

)2)(2(lim

2

4lim

2

2

2 −

+−=

−

−→→ x

xx

x

x

xx= 4

Verbalise: the 0 over 0 indeterminate from – we find the limit using an algebraic trick.

Sometimes, limits can be found using analytical considerations and sometimes, using the Table, Rules and a Decision Tree for Limits (practically the same as the one for sequences). 14.1.4 Applications of the Concept of a Limit

1. Imagine a person walking over a landscape represented by the graph of y = f(x). Her horizontal position is measured by the value of x, much like the position given by a map of the land or by a global positioning system. Her altitude is given by the coordinate y. She is walking towards the horizontal position given by x = a. As she does so, she notices that her altitude approaches L. If later asked to guess the altitude over x = a, she would then answer L, even if she had never actually reached that position. http://en.wikipedia.org/wiki/Limit_of_a_function#Motivation 2. It is often important for engineers to assess how a measured quantity (current, voltage, density, load) behaves with time: does it grow without bounds (then its limit is ∞ ), tends to a specific value (then its limit is a finite number), oscillates (then it has no limit) etc. 14.1.5 Existence of a Limit of a Function

A limit of a function does not always exist.

Examples:

1. xx

sinlim∞→

does not exist.

2. xx

1sinlim

∞→= 0.

14.2 Continuity of a function

A function f (x) is said to be continuous at point a if ax

lim→

f (x) = f (a).

Examples: where the filled point belongs to the graph and an empty circle – does not.

0

0

a x

y = f(x) (continuous at a) f(a)

y = f(x) (discontinuous at a)

a x

20

If a function is continuous at each point of an interval it is continuous on this interval. Intuitively, a continuous function is a function for which, smaller and smaller changes in the input (argument or independent variable) result in smaller and smaller changes in the output (value of a function or dependent variable). Otherwise, a function is said to be discontinuous. If f1 (x) and f2 (x) are continuous on an interval, so is f1 (x) f2 (x) (provided f2 (x) ≠ 0 when dividing). 14.2.1 Applications of Continuous Functions

“An analogue signal is a continuous signal for which the time varying feature of the signal is a representation of some other time varying quantity... For example, in sound recording, fluctuations in air pressure (that is to say, sound) strike the diaphragm of a microphone which causes corresponding fluctuations in a voltage or the current in an electric circuit. The voltage or the current is said to be an ‘analog’ of the sound.” http://en.wikipedia.org/wiki/Analog_signal 14.3 Differentiation of a real function of one real variable Differentiation is the second advanced operation on functions we cover. The result is a function, which is called the derivative of the original function. 14.3.1 A Derivative of a Function

A continuous function f (x) is differentiable at x = a if there exists x

afxaf

x ∆

−∆+

→∆

)()(lim

0=

= ).(')()(tantanlimlim 000

afadx

dfa

dx

dy

x

y

xx≡≡===

∆

∆

→∆→∆αα This limit is called a derivative.

Thus, we arrived at the following definition of the derivative:

x

xfxxf

dx

xdf

x ∆

−∆+=

→∆

)()(lim

)(

0.

+ -

× ÷

tangent line

α0

a

y = f (x)

x

α

a a + ∆x x

y = f (x)

f (a + ∆x)

f (a) ∆y

∆xx

0

0

21

14.3.2 Geometrical Interpretation of a Derivative

A derivative of a function is a local slope (gradient) of the function at a point. More specifically, it is the slope of the line tangent to the graph of the function at this point. 14.3.3 Mechanical Interpretation of a Derivative

A derivative of a function is a local rate (speed) of change of the function at a point. 14.3.4 Existence of a Derivative

A derivative of a function does not always exist. Here are graphs of functions which have no derivative at a point a in their domain: Examples:

1. Differentiate y = c (constant)

a) Can differentiate using a graph: the slope is the same everywhere and is zero. b) Can differentiate using the basic definition:

0lim)()(

lim00

=∆

−=

∆

−∆+

→∆→∆ x

cc

x

xfxxf

xx

2. Differentiate y = c x

y

a) Can differentiate using a graph: the slope is the same everywhere – c b) Can differentiate using the basic definition:

a x

y

a x

y

x

α0

cx

cxxxc

x

xfxxf

xx=

∆

−∆+=

∆

−∆+

→∆→∆

)(lim

)()(lim

00

x

y

1

0=⇒dx

dy Answer:

cdx

dy=⇒Answer:

22

14.4 A historical note

“When Newton and Leibniz first published their results on calculus in the 17th century, there was great controversy over which mathematician (and therefore which country) deserved credit. Newton derived his results first, but Leibniz published first. Newton claimed Leibniz stole ideas from his unpublished notes, which Newton had shared with a few members of the Royal Society. This controversy divided English-speaking mathematicians from continental mathematicians for many years, to the detriment of English mathematics. A careful examination of the papers of Leibniz and Newton shows that they arrived at their results independently, with Leibniz starting first with integration and Newton with differentiation. Today, both Newton and Leibniz are given credit for developing calculus independently. It is Leibniz, however, who gave the new discipline its name. Newton called his calculus ‘the science of fluxions’ ". http://en.wikipedia.org/wiki/Calculus


• Revise ALGEBRA Summary (particularly, the words term, sum, factor, product)




• Attempt the following exercises: Q1. Find

a) 20

cos1lim

x

x

x

−

→ (using 1

sinlim

0=

→ x

x

x and trigonometric identities)

b) 2

eelim

xx

0

−

→

−

x

c) 3

2

0 x

3xsinlim→x

d) 32x

1xlim

2

2

0 +

−

→x

e) 32x

1xlim

2

2

+

−

∞→x

Q2. Discuss continuity of y = tan x. Q3. Plot y = e

x. Differentiate this function approximately using the graph and plot the

approximate derivative .xe

dx

d

Q4. Prove

a) 1−= nn nxxdx

d (Hint: use the Binomial Theorem – look it up in books or Google it.)

23

b) xx eedx

d=

c) x xdx

dcossin =

d) sin x xcosdx

d−=

24

Lecture 15. DIFFERENTIAL CALCULUS: Differentiation (ctd.) Using the first principles (that is, the definition of a derivative) we can create a table of derivatives of elementary functions. 15.1 Differentiation Table

Examples:

1. =dx

πcos d0

2. 23

3d

dt

t

t=

y = f(x) dx

xdf )(

constant

0 This row means 0

const =

dx

d

xn

nxn-1

This row means 1-nn

nx x

=dx

d

ex

ex This row means x

x

edx

d e=

ln x

x

1 This row means

xdx

d 1ln x =

sin x

cos x This row means x

dx

dcos

sin x =

cos x

− sin x This row means x

dx

dsin

x cos −=

Differentiation variable? x Function to differentiate? constant It is a table function

Differentiation variable? t Function to differentiate? power 3

it is the It is a table function of t

function to differentiate

argument differentiation variable

25

3. vdv

vd cos

sin=

4. 23

)(ln3)(ln

)(lnt

td

td=

5. v

v

v

ede

edsin

cos−=

To differentiate combinations of elementary functions we can use Differentiation Table, Differentiation Rules and Decision Tree given in figure 15.1. 15.2 Differentiation Rules Note: These rules can be proven using the definition of the derivative as a limit and rules for limits.

1. dx

xdf

dx

xfd )()(α

α= - constant factor out rule.

2. dx

xdg

dx

xdf

dx

xgxfd )()()]()([+=

+ – sum rule.

3. )()(

)()())()((

xfdx

xdgxg

dx

xdf

dx

xgxfd+= – product rule.

4. 2

)()(

)()(

)(

)(

g

xfdx

xdgxg

dx

xdf

dx

xg

xfd −

= – quotient rule.

5. dx

dg

dg

gdf

dx

xgfd )())(((= – chain rule (decompose, differentiate, multiply). Here

g = g (x).

linearity property

Differentiation variable? v Function to differentiate? sin( )

It is a table function of v

Differentiation variable? ln t Function to differentiate? power 3

it is the It is a table function of ln t

Differentiation variable? ve

Function to differentiate? cos( )

it It is a table function of ve

26

Examples:

1. xdx

x d cos2

sin2=

2. xxdx

x) x d (sin - cos

cossin=

+

3. 322323 22

+=++=++

xdx

d

dx

x d

dx

d x

dx

)x d (x

Differentiation variable? x Function to differentiate? xsin 2 Last operation in this function? Multiplication Is one of the factors constant? Yes Apply the “constant factor out” rule

Constant factor α = 2

Variable Factor )(xf = xsin , xdx

x d

dx

d f(x) cos

sin==

Differentiation variable? x

Function to differentiate? 2 3 2 ++ xx Last operation in this function? Addition Apply the sum rule Go over the same reasoning for each term

Differentiation variable? x Function to differentiate? xx cos 5 3sin + Last operation in this function? Addition Apply the sum rule

f-line: 1st term f(x) = xsin , xdx

x d

dx

d f(x) cos

sin==

g-line: 2nd term g(x) = cos x, xdx

x d

dx

d g(x) sin

cos−==

27

4. xx xdx

x)d (x cossin

sin+=

x

5. 1lnln

+= xdx

x)d (x (using the product rule)

6. 2

sincos

sin

x

x x - x

dx

)x

xd (

= (using the quotient rule)

Last operation in this function? Multiplication Is one of the factors constant? No Apply the product rule

f-line: 1st factor f(x) = x, 1==dx

d x

dx

d f(x)

g-line: 2nd factor g(x) = sin x , xdx

x d

dx

d g(x) cos

sin==

f-line: 1st factor f(x) = x, 1)(

==dx

dx

dx

xdf

g-line: 2nd factor g(x) = ln x , xdx

xd

dx

xdg 1ln)(==

f-line: Numerator f(x) = sin x, xdx

x d

dx

d f(x) cos

sin==

g-line: Denominator g(x )= x, 1==dx

dx

dx

d g(x)

28

7. 22

ln1ln

1ln

x

x

x

x - x

x

dx

)x

xd (

−=

⋅=

8. dx

x d 2sin = 2 cos 2 x

Differentiation variable? x

f(x) ? x

xln

Last operation in this function? Division Apply which rule? Quotient rule

f-line: Numerator f(x)=ln x, xdx

x d

dx

d f(x) 1ln==

g-line: Denominator g(x)=x, 1==dx

dx

dx

d g(x)

Differentiation variable? x f(x)? x 2sin Last operation in this function? sin( ) Its argument? 2 x Apply which rule? The chain rule

f-line: Last operation f( )= sin( ), xgdg

(g) d

dx

d f(x) 2coscos

sin===

g-line: Its argument g(x)=2x, 22

==dx

x d

dx

d g(x)

29

9. dx

x d 3ln =

xx

1

3

13 =

15.3 Decision Tree for Differentiation Generally, to differentiate a function we use the Decision Tree for Differentiation (see figure 15.1 below) that allows us to decide whether to use the Differentiation Table (containing derivatives of elementary functions) straight away or first use Differentiation Rules (on how to differentiate combinations of elementary functions). Once the rule is applied, even if the problem is not solved, it is reduced to one or several simpler problems which have to be treated using the Decision Tree again. To use the Decision Tree start at the top and follow the arrows which are associated with the correct answers (if any). Use it as a formula, substituting your differentiation variable for x and your function for f(x).

15.4 The higher order derivatives If a derivative of f(x) is a differentiable function its derivative is called the second derivative of f(x)

If the second derivative of f(x) is differentiable its derivative is called the third derivative of f(x),

)(2

2

3

3

dx

fd

dx

d

dx

fd= etc.

Differentiation variable? x Function to differentiate? x3ln Last operation in this function? ln( ) Its argument? 3 x Apply the chain rule

f-line: Last operation f( ) = ln ( ), x

g

dg

(g) d

dx

d f(x)

3

11ln===

g-line: Its argument g(x) = 3x, 33

==dx

x d

dx

d g(x)

)(2

2

dx

df

dx

d

dx

fd=

30

Figure 15.1. Decision Tree for Differentiation. 15.5 The partial derivatives If we consider a function of say, two variables (two arguments) f(x,y) its partial derivatives with

respect to x and y are denoted by x

yxf

∂

∂ ),( and

y

yxf

∂

∂ ),(, respectively.

15.6 Applications of differentiation Later you will find that all simple engineering systems that you are going to study are described by differential equations that relate values of measured variables, such as displacement, angle or rotation, concentration, current or voltage to their first or second derivatives. You will also learn that derivatives and limits are helpful in sketching functions and thus illustrating or even predicting behaviour of the above measurable variables. Finally you will find out how

x

÷

Maybe, tricks Maybe, tricks

No Yes Yes No


Is y(x) composite?

What is the last operation in y(x)?

No

The answer is

in the table

Is y(x) a constant or another table function of x?

Product rule or

quotient rule

Constant factor

outside Chain rule Use tricks

Yes

Other ±

?)(xydx

d

?

What is the differentiation variable ? (it appears after d at the bottom)

What is the function to differentiate ? (it appears after d at the top)

Sum rule

31

derivatives can help to approximate functions, e.g. how your calculators calculate sin, cos values etc. 15.7 Instructions for self-study


• Revise Summaries on the ORDER OF OPERATIONS and FUNCTIONS

• Revise Lecture 4 and Solutions to Exercises in Lecture 4 (particularly, the operation of composition)




• Attempt the following exercises:

Q1. Find a) 1

2

2

1

+

+

te

td

d

b) 1

2

2

1

+

+

xe

td

d

Q2. Find the derivatives

a) uudu

dsin)1( 2 +

b)u

u

du

d

sin

12 +

Q3. Find the derivatives

a) dx

xd 1cos 2 +

b) dy

xd 1cos 2 +

Q4. Differentiate y with respect to t if y = .)( 22 natt ++

32

Lecture 16. DIFFERENTIAL CALCUUS: Sketching Graphs Using Analysis In general, any composite function can be sketched using our knowledge of limits and derivatives. In this Lecture we show how. 16.1 Stationary points A stationary point is a point x0 in the function domain where its derivative f’(x0) = 0. Examples: Sketch the following functions and find their stationary points x0 : 1. y = x

2 2. y = x

3 3. y = e

x 4. y = sin x

0=dx

dy 0=

dx

dy 0≠= x

edx

dy 0cos == x

dx

dy

2x = 0 3x2

= 0 No stationary points x0 = ,2

nπ

x0 = 0 x0 = 0 16.2 Increasing and decreasing functions If derivative (local slope) f’ (x) > 0 on interval I, function f (x) is said to be increasing on I. If derivative (local slope) f’ (x) > 0 on interval I, function f(x) is said to be decreasing on I.

n – odd integer (n = 2 m + 1, m – integer)

x

y

y

x

x0 x x0 x x0 x x0 x

f (x) f (x) f (x) f (x)

y = ex

x

y = x2

x0 x x0 x

y = x3

2

π

2

3π x

y = sin x

33

Examples: Functions y = ex and y = ln x are increasing everywhere; functions y = │x│

and y = x2 are decreasing for x < 0 and increasing for x > 0.

16.3 Maxima and minima

If derivative (local slope) f’ (x) > 0 to the left of x0 and function f’ (x) < 0 to the right of x0, then function f (x0) is said to have a maximum at x0.

If derivative (local slope) f’ (x) < 0 to the left of x0 and function f ’(x) > 0 to the right of x0, then function f (x0) is said to have a minimum at x0. Examples: y = x

2 has minimum at 0, x3 has no minimum or maximum, sin x has

maxima at x = m2

π when m is odd and minima when m is even.

Examples: 1. Sketch y = x

2 + 3x +1 by completing the square

Solution:

Completing the square, y = (x + 2

3)

2 -

4

9 + 1 = (x +

2

3)

2 -

4

5.

Step 1. Using the recipe for simple transformations, we first drop all constant factors and terms and sketch the basic shape of the function that remains: Step 2. We then reintroduce the constant factors and terms that appeared in the original functional equation one by one, in Order of Operations and sketch the resulting functions underneath one another:

y = x2

x

0 x -

2

3

y = (x +2

3)2

+2

3 – constant term, first operation

⇒ translation wrt x-axis by -2

3

x0 x y

y

x0 x

34

Step 3. Intersection with the y-axis: x = 0, y = 4

Intersection with the x-axis: y = 0, x1, 2 =2

1,

2

12

2

1.23

2

53

2

493−−≈

±−≈

±−=

−±−

2. Sketch y = x

2 + 3x +1 by analysis.

Solution Step 1. Intersection with the y-axis: x = 0, y = 4

Intersection with the x-axis: y = 0, x1, 2 = 2

1.23

2

53

2

493 ±−≈

±−=

−±−

Step 2. Find and sketch the first derivative y’ = 2x + 3

a) Find where the first derivative is zero: 2x + 3 = 0 ⇒ x = −2

3

⇒Stationary point of the function is x = −2

3

b) Find where the first derivative is positive and where it is negative: Can do so by solving the inequalities

2x + 3 > 0 (y’ > 0) for x > −2

3

2x + 3 < 0 (y’ < 0) for x < −2

3

Note: inequalities can be solved the same way as equations but when multiplying by negative factors the inequality sign should be reversed.

-4

5 – constant term, first operation

⇒ translation wrt y-axis by -4

5

y = (x + 2

3)

2 -

4

5

4

2

53 −− -

2

3 --

2

53 +

4

5− x

35

or graphically Step 3: Sketch the function y directly underneath

the first derivative y’ Step 4: Find the minimum function value

4

5

4

41891

2

9

4

9)

2

3( −=

+−=+−=−y

y = x

2 + 3x +1

decreasing increasing

y’ = 2x + 3

Ne negative -2

3 positive

x

-4

5 x

2

53+− -

2

53 − -

2

3

36

3. Sketch y = x

3 + 3x

2 + 4x + 1 by analysis

Solution Step 1. Find and sketch the first derivative y’ y’ = 3x

2 + 6x + 4 = 3 (x

2 + 2x) + 4 = 3 [(x + 1)

2 - 1] + 4 = 3 (x + 1)

2 + 1 > 0

y’ > 0 everywhere Step 2. Sketch the function y underneath y is increasing everywhere Special points: y(0) = 1 y( − 1) = − 1 (not that special but easy to find)

y’

-1

1

x

1

-1

y

-x -1

37

Optional 16.4 Convex and concave functions

If the second derivative f’’ (x) > 0 on interval I

the first derivative f’ (x) is increasing on I and function f (x) is said to be convex on interval I. If the second derivative f’’(x) < 0 on interval I

the first derivative f’ (x) is decreasing on I and function f(x) is said to be concave on interval I. Examples: establish the regions of convexity of the following elementary functions:

2 xy = 3 xy = y = e x

y’ = 2x y’ = 3 x2 y’ = e

x

y’’’ = 2 > 0 > 0 if x > 0 y’’ = e x > 0

xy 6"=

⇒ 2 xy = is convex < 0 if x < 0 ⇒ y = ex is convex

everywhere convex if x > 0 everywhere

⇒ 3 xy = is

concave if x < 0

16.5 Inflexion points

If the second derivative f’’(x0) = 0 and f’’(x) changes sign at the stationary point x0, then the stationary point x0 is called an inflexion point. Example: y = x

3 has an inflexion point at x = 0. Thus, there is another way of determining whether the stationary point is at a maximum or minimum:

If f’ (x0) = 0 and f’’(x0) > 0, then the function is convex in the vicinity of x0 and x0 is at a

minimum.

If f’(x0) = 0 and f’’ (x0) < 0, then the function is concave in the vicinity of x0 and x0 is at a maximum.

If f’ (x0) = 0 and f’’(x0) = 0, then further checks are required.

x

y

x

y

38

16.6 Sketching rational functions using analysis We illustrate the general principles of sketching by analysis by applying it to rational functions.

Examples: 1. Sketch the rational function y = 1

1

+

−

x

x.

Solution Step 1. Domain: x ≠− 1 (x = − 1 is called a pole, the line x = − 1 is called a vertical asymptote) Step 2. Intersection with the y-axis: x = 0 , y = − 1 Intersection with the x-axis: y = 0 , x = 1 Step 3. Behaviour at the domain boundaries: x → ∞, y → 1

x → -∞ , y → 1 Step 4. Find the first derivative:

0)1(

2

)1(

)1(1

22>

+=

+

−−+=

xx

xx

dx

dy

⇒ y is increasing everywhere

A rational function R(x) is

R(x) = )x(Q

)x(P, where P and Q are polynomials.

The line y=1 is called a horizontal asymptote. We

know that the function approaches it when x grows

large but we do not know how.

-1 x

y

y

1 -1 x -1

? ?

1

y

1 -1 -1

? ?

-1

increasing increasing 1

1 -1 x

x

y

39

2. Sketch y = 44

1

2 +−

−

xx

x.

Solution Step 1. Domain: x ≠ 2 (the x=2 value is called a pole, line x=2 - a vertical asymptote)

Step 2. Intersection with the y-axis: x = 0, y = −4

1

Intersection with the x-axis: y = 0, x = 1 Step 3. Behaviour at the domain boundaries x → ±∞, y → ±0

The line y = 0 is a horizontal asymptote Step 4. Find the first derivative:

3)2( −

−=x

x

dx

dy

4

2

)2(

2

−

+−=

x

xx

-ve +ve -ve decr incr decr

Note that x = 2 turns the numerator xx 22 +− into zero but dx

dy is not zero at this

point. The “phantom” zero appears because we multiplied both numerator

and denominator of dx

dy by (x − 2). This was a trick performed to make sure that

the denominator is never negative.

dx

dy ~ − x

2 + 2x

2

0= at x = 0

> 0 (y is increasing) for 0 < x < 2 < 0 (y is decreasing) for x < 0 & x > 2

This is not a derivative but it is 0,

positive and negative where dx

dyis

– since 4)2( −x >0 when 2≠x .

The sign ~ means “behaves as”.

x

4

1−

y

x

40

Optional

Sketch y = 1x

2x4x 2

+

++ - improper rational function

Difference in degrees = 1 ⇒ the whole part is a polynomial degree 1

1

242

+

++

x

xx = Ax+B +

1+x

C

To find A, B and C use

1. long division or 2. partial fractions method

y = x + 3 − 1x

1

+

Now we have to sketch y = x + 3 −1x

1

+

Step 1. Domain: x ≠ − 1 (point x = − 1 is a pole and line x = − 1 is a vertical asymptote) Step 2. Intersection with the y-axis: x = 0, y = 2

Intersection with the x-axis: y = 0, x2 + 4x + 2 = 0, x1,2 = − 2 2±

Step 3. Behaviour at the domain boundaries: x → ±∞, y → x + 3 – an oblique asymptote (typical for improper rational fractions)

polynomial degree 2

polynomial degree 1

1

242

+

++

x

xx = Ax + B +

1+x

C (x + 1)

x

2 + 4x + 2 = Ax(x + 1) + B(x + 1) + C

True for all x ⇒ can choose any convenient value of x 1. Choose values that “kill” terms x = − 1: 1 – 4 + 2 = C, C = − 1

x = 0: 2 = B + C, 2 = B – 1, B = 3

2. Equate coefficients of the highest power: 1 = A

polynomial degree 1

polynomial degree 0

41

Step 4. Find the first derivative:

2

2

)1(

22

+

++=

x

xx

dx

dy> 0, because x2

+ 2x + 2 = 0 at x1,2 = − 1 j± (no real roots)

16.7 Applications of rational functions and their graphs In control theory, single input single output (SISO) linear dynamic systems are often characterised by the so-called transfer functions. Any such transfer function is a rational function of one complex variable. Sketching magnitudes and phases of these functions helps design efficient control systems. 16.8 Instructions for self-study

• Revise Lecture 9 and study Solutions to Exercises in Lecture 9 (sketching by simple transformations)

• Revise ALGEBRA Summary (addition and multiplication, factorising and smile rule, flip rule)


• Revise Lectures 13 - 15 (limits and differentiation) and study Solutions to Exercises in Lecture 14 using the STUDY SKILLS Appendix

• Study Lecture 16 using the STUDY SKILLS Summary

• Do the following exercises: Q1. Sketch using analysis a) y = − 3 (x

3 − 3x

2 + x)

-1

2

x

dx

dy ~ x2

+ 2x + 2

positive everywhere

y = x + 3

-2- 2

-2+ 2

y

x

increasing everywhere

42

b) y = −8

1

4

1

2

1 2 ++ xx

Q2. Sketch y = 1

65

2

2

+

+−

x

xx

Q3. Sketch y = u(x + 2) 1

652

2

+

+−

x

xx

Q4. Sketch y = 1

652

+

+−

x

xx

43

Lecture 17. Application of DIFFERENTIAL CALCULUS to Approximation of Functions: the Taylor and Maclaurin Series 17.1 Approximating a real function of a real variable using its first derivative Let us discuss how we can approximate the value of function f(x) at a+∆x if we know its value at a. It is clear from the figure below that f (a + ∆x) ≈ f (a) + y∆ .

How can we approximate y∆ ? To see this, let us revise the definition of the first derivative

given in Lecture 13.

⇒ f (a + ∆x) ≈ f (a) + f’’(a) ∆x Verbalise: The function value near a approximately equals the function value at a + derivative at a times distance to a. This is an approximation linear in ∆x (in the vicinity of x the graph of f(x) looks like a straight line). Sometimes we need approximations based on quadratic, cubic … n-th degree polynomials. Before showing how to find such approximations we introduce a new way of representing a polynomial. 17.2 The Maclaurin polynomials

Question: What is a polynomial? Answer:

Question: Conventionally we write the first term of the polynomial )(xPn as nnxa . All

coefficients are constant with respect to x and are described by the letter a. What will be the next term and the next and the next…? Answer: Any polynomial can be re-written in another form as

the Maclaurin polynomial: n(n)nnn

nnn xn!

)(P...x

!

)(Px

!

)(P)x(P)(P(x)P

0

3

0

2

000 3

'''2 ++

′′+′+=

where n factorial nn ×××××= ...4321! .

Question: What is the geometrical interpretation of derivative? Answer: Question: How can we find its approximate value? Answer: Question: How to solve this approximate equality for y∆ ?

Answer:

α

a a + ∆x x

y = f (x)

f (a + ∆x)

f (a) y∆

x∆ x

44

We can show this using the following steps:

01

12

23

31

1 ....)( axaxaxaxaxaxPn

nn

nn ++++++= −− ⇒ nP (0) = 0a ⇒ 0a = nP (0)

122

32

11

23....)1()( axaxaxnanxaxPn

nn

nn ++++−+=′ −−

− ⇒ )0(nP′ = 1a ⇒ 1a = )0(nP′

233

12

223....)2)(1()1()( axaxnnaxnnaxPn

nn

nn +⋅⋅++−−+−=′′ −−

− ⇒ )0(nP ′′ =2 2a ⇒ 2a =2

)0(nP ′′

33

23....)2)(1()( axnnnaxPn

nn ⋅⋅++−−=′′′ − ⇒ )0(nP ′′′ = 332 a⋅ ⇒ 3a = !3

)0( nP ′′′

… …

nnn

nn xnnnnnaxP

−−−−−= )]1()....[2)(1()()(

⇒ )0()(n

nP =n! na ⇒ na = !

)0()(

n

Pn

n

⇒ n(n)nnn

nnn xn!

)(P...x

!

)(Px

!

)(P)x(P)(P(x)P

0

3

0

2

000 32 +

′′′+

′′+′+=

Verbalise: The Maclaurin polynomial at x is the polynomial at 0 + 1st derivative at 0 times x to 0 + 2nd derivative at 0 times x squared over 2! + …. 17.3 The Taylor polynomials In the same way we can prove that any polynomial can be re-written as

Verbalise: The Taylor polynomial at x is the polynomial at a + derivative at a times the signed distance from x to a + …. Note: in the Maclaurin series x can be viewed as the signed distance from x to 0. 17.4 The Taylor series Many differentiable functions f(x) can be represented using the infinite

Verbalise: The function value at x is the function value at a + derivative at a times the signed distance to a + ….

Taylor series: ...)2(!2

)("))((')()( 2 +−+−+= x

afaxafafxf

the Taylor polynomial: n(n)n

nnn a)(xn!

(a)P...a)(a)(xP(a)P(x)P −+−′+=

45

These functions can be approximated as polynomials using truncated Taylor series (that is, Taylor polynomials).

17.5 The Maclaurin series If a = 0, the Taylor series is called the Maclaurin series. Examples:

1. Find the Maclaurin series for ex.

Solution Step 1. The general Maclaurin series is

...!2

)0(")0(')0()( 2 +++= x

fxffxf

,

Step 4. Substitute the above values into the general the Maclarin series:

!...

!3!21

32

n

xxxxe

nx +++++= …

⇒ 11 <+≈ xx,x

e (in the vicinity of 0 the graph of ex is almost a straight line)

Step 2. Function and its derivatives are

Step 3. Values of the function and its derivative at 0 are

xexf =)(

xexf =)('

xexf =)("

…

xn exf =)()(

1)0( =f

1)0(' =f

1)0(" =f

...

1)0()( =nf

Consider the approximation error Optional

),()(),( xaPxfxae nn −≡

If the error en (x) → 0 as n → ∞, then f (x) equals its Taylor series at x and y(x) ≈≈≈≈ Pn (a, x) is an approximation based on the n-th order Taylor polynomial.

For ex, sin x, cos x, 0))(,( →− axxae n faster than nax )( − .

Proving the above results and checking the limiting behaviour of the error ne is

rather involved.

46

2. Find the Maclaurin series for sin x.

Solution Step 1. The general Maclaurin series is

...!2

)0(")0(')0()( 2 +++= x

fxffxf



xxf cos)(' =

xxf sin)(" −=

xxf cos)( −=′′′

xxfIV sin)()( =

…

1)0('f =

0)0("f =

1)0( −=′′′f

0)0()( =IVf

…

Step 4. Substitute the above values into the general Maclarin series:

...!7!5!3

sin753

+−+−=xxx

xx

⇒ 1,sin <≈ xxx (in the vicinity of 0 the graph of sin x is almost a straight

line) c) Find the Maclaurin series for cos x.

Solution:

Step 1. ...!2

)0(")0(')0()( 2 +++= x

fxffxf



xxf cos)( =

xxf sin)(' −=

xxf cos)(" −=

…

1)0( =f

0)0(' =f

1)0(" −=f

… Step 4.

...!4

x

!2

x1xcos

42

++−= -

⇒ 12

1cos2

<−≈ x, x

x (in the vicinity of 0 the graph of cos x looks like a

parabola – reflected wrt (with respect to) the horizontal axis shifted up by 1).

47

Thus, while any polynomial can be re-written as a Taylor or MacLaurin polynomial, many other n time differentiable functions can be approximated by Taylor or MacLaurin

polynomials (x)Pn . The discussion on whether such approximation is possible or how to

decide the optimal degree of the approximating polynomial lies outside the scope of these notes. 17.6 L’Hospital’s rule

Let us find)(

)(lim

xg

xf

ax→ when 0)(lim)(lim ==

→→xgxf

axax, so that we have the 0/0 indeterminate

form. If both )(xf and )(xg have Taylor expansion in the vicinity of a we can write

)('

)('lim

...))((

...))((lim

)(

)(lim

xg

xf

axag

axaf

xg

xf

axaxax →→→=

+−′

+−′= .

The recipe

)('

)('lim

)(

)(lim

xg

xf

xg

xf

axax →→= if 0)(lim)(lim ==

→→xgxf

axax

is called L’Hospital’s rule. If still get the 0/0 indeterminate form – continue applying the rule until arrive at a determinate answer (find the second derivatives of the numerator and denominator, their third derivatives etc.) L’Hospital’s rule can be extended to cover functions with just one derivative, +∞=a or

∞− and ∞∞ / indeterminate forms but these proofs lie outside the scope of these notes.

Examples:

1) Find x

x

x

sinlim

0→.

Solution: 0sinlimlim00

==→→

xxxx

. Applying L’Hospital’s rule,

2) Find x

ex

x

−

→

1lim

0.

Solution: .01limlim00

=−=→→

x

xxex Applying L’Hospital’s rule, 1

1lim

1lim

00−=

−=

−

→→

x

x

x

x

e

x

e

17.7 Applications The above series are used by calculators to give approximate values of elementary functions, sin, cos, exponent etc.


11

coslim

sinlim

00==

→→

x

x

x

xx

:(x - a)

48

• Revise ALGEBRA Summary (smile rule)

• Revise Lecture 5 (polynomials)

• Revise FUNCTIONS Summary

• Revise Lectures 13 - 14 (limits)

• Revise Lectures 14 - 15 (differentiation) and Solutions to Exercises in Lecture 14 using the STUDY SKILLS Appendix



• Do the following exercises: Q1.

a) Find the first three non-zero terms in the Maclaurin series forx

y+

=1

1

b) Use the first two terms of the above Maclaurin series to approximate x

y+

=1

1in the

vicinity of x = 0

c) Find, without using a calculator, the approximate value of 5

4

d) Find )]1(1

1[lim 1

1

x

xe

x

+

−→−

+.

Q2.

a) Find the first three non-zero terms in the Maclaurin series forx

y−

=1

1

b) Use the first two terms of the above Maclaurin series to approximate x

y−

=1

1in

the vicinity of x = 0


4

d) Find )]1sin(1

1[lim

1x

xx−

−→

Q3.

a) Find the first three non-zero terms in the Maclaurin series for xy += 1

b) Use the first two terms of the above Maclaurin series to approximate xy += 1 in

the vicinity of x = 0


4

d) Find 21 1

1lim

x

x

x −

+

−→

Q4.

a) Find the first three non-zero terms in the Maclaurin series for xy −= 1

b) Use the first two terms of the above Maclaurin series to approximate xy −= 1 in the

vicinity of x = 0


4

49

d) Find 211

1lim

x

x

x −

−

→

50

Lecture 18. INTEGRAL CALCULUS: Integration of Real Functions of One Real Variable (Definite Integrals)

Integration (over a finite interval) of a function is a new advanced operation on functions. If it exists, the result is a number (or a constant with respect to the independent variable) called a definite integral. 18.1 A definite integral Consider the area A between a curve, which is a graph of a function f(x), the horizontal axis and lines x = a and x = b:

Question: What is the area of first rectangle, second rectangle etc.? Answer:

In other words, this area A approximately equals

∑−

=− ∆=∆++∆+∆+∆≈

1

0

1210 )()(...)()()(n

i

in xxfxxfxxfxxfxxfA

where ax =0 and .bxn =

Question: How many rectangles do we have here? Answer: Taking the limit ,0→∆x so that the width of the rectangles gets smaller and smaller, the

number of rectangles grows larger and larger ( ∞→n ) and we can write that this area A exactly equals

∑ ∫−

=→∆≡∆=

1

00)()(lim

n

i

b

a

ix

dxxfxxfA

Thus, we introduced the definition of the definite integral of a function )(xf over the interval [a, b],

∑∫−

=→∆∆=

1

00)(lim)(

n

i

i

b

ax

xxfdxxf

The geometrical interpretation of the definite integral of a real function of one real variable is a “(signed) area between a curve (the graph of this function) and horizontal

Area A between the curve and horizontal axis approximately equals the sum of areas of inscribed thin rectangles

if the limit exists

a b x a x1 x2 … b x

y = f(x) y = f(x)

∆ x ∆ x

A=?

51

∫b

a

dx...

∫

axis”. The area is signed, because on the interval where )(xf is negative, all products

)( ixf x∆ are negative. Thus, when the horizontal axis is “above the curve” the integral is

negative and equal the area between the curve and horizontal axis with the – sign.

Example: ?sin2

0

=∫π

xdx Answer: … (sketch the graph to see that).

18.2 Notation

- integral (an elongated letter S reminds us that we are talking about the limit of a sum Σ )

- a definite integral between a and b (the notation reminds us that we are talking

about the limit of a sum of areas of thin rectangles of width x∆ fitted between the curve, horizontal axis and straight lines x = a and x = b). a, b are called the lower and upper limits of integration, respectively. 18.3 Comparison of a definite Integral and derivative

∆ y ∆ x Whether defining an integral or derivative we reduce a difficult problem involving a curve to many simple problems involving small straight segments. Ability to reduce complicated unfamiliar pictures to simple familiar patterns is an invaluable engineering skill. 18.4 Examples of integrable functions 1. Any f(x) continuous in [a, b] can be integrated Question: Why? Answer: 2. Any bounded f(x) with finite number of jumps can be integrated Question: Why? Answer: Note: If change f(x) at a finite number of points – this does not change integrability. Question: Why?

The derivative approximatlely equals the slope of local inscribed segment

≈dx

dy

x

y

∆

∆

x

y

a b x

y

y

∆ x x

y

x

52

),()( abdxxf

b

a

−=∫ µ

Answer:

Question: Is continuity sufficient for integrability, necessary or both? Answer: Note: Continuity is not sufficient for differentiability. Continuity is necessary for differentiability. 18.5 The Mean Value Theorem If f(x) is continuous, then there exists a value µ, such that m ≤ µ ≤ M and where m is the smallest value of f(x) on the interval [a, b] and M is the largest values of f(x) on the interval [a, b]. µ is the Greek letter mu. y = f(x)

Why is it called that?

1. The mean value of the sequence {1, 3} is 2

31+

2. The mean value of the sequence {x0, …, xn-1} is n

x n 1-10, x…x ++

Revision

Let C be a condition. Let E be an event.

If EC ⇒ , then the condition C is sufficient for event E to take place. If CE ⇒ , then the condition C is necessary for event E to take place.

µ = ∫−

b

a

dxxfab

)(1

is called the mean value of f(x) on the interval [a, b].

The area under the curve (between a and b) equals

)( ab −µ , the area of a rectangle (in red, long dashes)

m

M

a c b x

µ

53

∫=Φx

a

dttfx ,)()(

3. The mean value of the sequence {f(x0) ,…, f(xn-1)} is )(1

1

0

i

n

i

xfn∑−

=

Question: With reference to Section 18.1, given a, b and x∆ , what is n, the number of

rectangles of width x∆ used to approximate the area between f(x), the horizontal axis and lines x = a and x = b ? Answer:

Question: What is xnx

∆→∆ 0

lim ?

Answer: This means that we can write

∫∑∑−

= →∆

−

=−

→∆∆

=b

a

n

i xi

n

i

i dxxfab

xxfxn

xfn

)(1

)(1

)(1

1

0 0

1

0

This is why µ = ∫−

b

a

dxxfab

)(1

is called the mean value of )(xf on the interval [a, b].

18.6 A definite integral with a variable upper limit - function Φ(x) Let )(xf be integrable in [a, b]. We can introduce a new advanced operation on a function

– evaluating a definite integral with a variable upper limit. The result is a function of x, since if a is fixed for every x we have one and only one value for the signed area between the curve and horizontal axis. Question: Does this area depend on t? Answer: y=f(x)

a x b x

54

18.7 The Fundamental Theorem of Calculus

Attempt the proof yourself (see Q2.) 18.8 Applications of integration Later you will find that apart from being useful in finding areas and volumes, integration is necessary to solve differential equations that describe all engineering systems that you are going to study. 18.9 A historical note According to Wikipedia, “integration can be traced as far back as ancient Egypt, circa 1800 BC, with the Moscow Mathematical Papyrus demonstrating knowledge of a formula for the volume of a pyramidal frustum. The first documented systematic technique capable of determining integrals is the method of exhaustion of Eudoxus (circa 370 BC), which sought to find areas and volumes by breaking them up into an infinite number of shapes for which the area or volume was known. This method was further developed and employed by Archimedes and used to calculate areas for parabolas and an approximation to the area of a circle. Similar methods were independently developed in China around the 3rd century AD by Liu Hui, who used it to find the area of the circle. This method was later used in the 5th century by Zu Chongzhi to find the volume of a sphere. That same century, the Indian mathematician Aryabhata used a similar method in order to find the volume of a cube.

The next major step in integral calculus came in the 11th century, when the Iraqi mathematician, Ibn al-Haytham (known as Alhazen in Europe), devised what is now known as "Alhazen's problem", which leads to an equation of the fourth degree, in his Book of Optics. While solving this problem, he performed an integration in order to find the volume of a paraboloid. Using mathematical induction, he was able to generalize his result for the integrals of polynomials up to the fourth degree. He thus came close to finding a general formula for the integrals of polynomials, but he was not concerned with any polynomials higher than the fourth degree. Some ideas of integral calculus are also found in the Siddhanta Shiromani, a 13thcentury astronomy text by Indian mathematician Bhāskara II.

The next significant advances in integral calculus did not begin to appear until the 16th century. At this time the work of Cavalieri with his method of indivisibles, and work by Fermat, began to lay the foundations of modern calculus. Further steps were made in the early 17th century by Barrow and Torricelli, who provided the first hints of a connection between integration and differentiation.

The major advance in integration came in the 17th century with the independent discovery of the fundamental theorem of calculus by Newton and Leibniz. The theorem

The Fundamental Theorem of Calculus : If f(x) continuous, then )()( xfdttfdx

dx

a

=∫ .

55

demonstrates a connection between integration and differentiation. This connection, combined with the comparative ease of differentiation, can be exploited to calculate integrals. In particular, the fundamental theorem of calculus allows one to solve a much broader class of problems. Equal in importance is the comprehensive mathematical framework that both Newton and Leibniz developed. Given the name infinitesimal calculus, it allowed for precise analysis of functions within continuous domains. This framework eventually became modern Calculus, whose notation for integrals is drawn directly from the work of Leibniz.”


• Revise Lecture 16 and study Solutions to Exercise in Lecture 16 using the STUDY SKILLS Appendix



• Attempt the following exercises: Q1. Find using the first principles, that is using the definition of definite integral,

a) ∫1

0

2.0 dp

b) ∫1

0

5dq

c) ∫1

0

5.0 du

d) ∫1

0

sds

e) ∫1

0

3udu

f) ∫1

0

4vdv

Q2. (advanced). Prove that if f(x) is continuous, then )()( xfdttfdx

dx

a

=∫ (using the first

principles, that is using the definition of definite integral, and Mean Value Theorem). Describe this formula in words.

56

Lecture 19. INTEGRAL CALCULUS: Integration of Real Functions of One Real Variable (Indefinite Integrals) 19.1 The relationship between differentiation and integration According to the Fundamental Theorem of Calculus if )(xf is continuous we have

)()( xfdttfdx

dx

a

=∫ (19.1)

Question: What is the order of operations on )(tf in the left hand side above? (treat

“ dt∫ .... ” as bracketing the integrand).

Answer: Thus, differentiation of a continuous function undoes what integration does, that is, the differentiation is inverse to integration.

Let us now subdivide the [a, x] interval into small subintervals with ax =0 and xxn = :

cxfxfxfxfxfxfxfxfxf

xfxfxfxx

xfdttf

dt

d

nnnx

nx

n

i

i

x

x

a

+≡−=−++−+−

=∆++∆+∆=∆∆

∆=

−→∆

−→∆

−

=→∆∑∫

)()()()]}()([...)]()([)]()({[lim

(19.2) )](...)()([lim)(

lim)(

0112010

1100

1

00

Question: What is the order of operations on )(tf in the left land side above?

Answer: Thus, integration is not a perfect inverse to differentiation, the definite integral with the variable upper limit depends on the lower limit too.

Another way to see this is as follows: Examining the graph below and remembering that the integral can be interpreted as a signed area, we can write

y=f(t)

a a0 x b t

57

cdttfdttfcdttfdttfdttf

x

a

x

a

x

a

a

a

x

a

+=+=+= ∫∫∫∫∫000

0

)()()()()( (19.3)

Since ∫=0

)(

a

a

dttfc is constant with respect to x but depends on 0a , Eq. (19.1) implies

that

)()),(( 0 xfcaxdx

d=+Φ

where instead of )(xΦ (the notation used in the last Lecture), we use a more logical

notation ∫=Φx

a

dttfax

0

)(),( 0 (a definite integral with a variable upper limit x actually

depends on the lower limit a0 too).

This means that each function )(xf is associated with many definite integrals ),( axΦ . For

this reason, the integral ),( axΦ is called an antiderivative of f(x), rather than the inverse

derivative of f(x) (remember, an inverse function – as any function – is supposed to assign only one value to any value of its argument). 19.2 An indefinite integral We have shown that for every continuous function there are infinitely many antiderivatives, differing from one another by a constant term. Often instead of the antiderivative we use a somewhat confusing shorthand

In view of (19.3), we have

cdttfdttf

x

a

x

a

+= ∫∫0

)()(

Substituting the above decomposition into definition (19.4) we get

This is a reiteration of the fact that the antiderivative is defined up to a constant term, because we do not know (and do not need to know) what is its lower limit of integration. We can now introduce a very important formula for evaluating definite integrals:

∫ dxxf )( ≡ ∫ +x

a

cdttf

0

)(

the indefinite integral ∫∫ ≡≡x

a

dttfdxxfxF )()()( (19.4)

(19.5)

58

This formula implies that

1. ∫∫ −=a

b

b

a

dxxfdxxf )()( (Why?)

2. if we know an indefinite integral we can evaluate (find the value of) any associated definite integral (note that the c-terms in F(b) and F(a) in Eq. (19.6) cancel out). Finally, note that using the indefinite integral notation the two Fundamental Theorems of Calculus (19.1) and (19.2) can be re-written as

Terminology

)(xf – integrand, x - integration variable, ∫ dx... is an indefinite integral with respect to x. It

has limits but these limits are allowed to vary, that is, are indefinite. 19.3 Finding an indefinite integral To find an indefinite integral we use the Integration Table of elementary functions, Rules (or Methods) for integrating combinations of functions and Decision Tree to decide which rule or Table entry to use. 19.3.1 The Integration Table

We can derive Integration Table for elementary functions using Differentiation Table. Differentiation Table Integration Table

g (x)

dx

xdg )(

constant 0

ex e

x

sin x cos x

cos x − sin x

xk

kxk-1

ln x, x>0

0,

1≠x

x

)()( xfdxxfdx

d=∫ , cxfdx

dx

xdf+=∫ )(

)(

)()()( aFbFb

a

dxxf −=∫

)(xf ∫= dxxfxF )()(

0 c (constant)

ex e

x +c

cos x sin x +c

sin x − cos x +c

xn, n≠-1

cn

xn

++

+

1

1

0,1

≠xx

ln│x│+c

⇒

(19.6)

(19.7)

integrand

integration variable

59

Let us check that xdxx

ln1

=∫ , for any x ≠ 0. Optional

Indeed, x , 0≥x

− x , 0<x

x

1, 0>x ,

)1

(x

−− , 0<x , where we used the chain rule of differentiation.

Examples:

1. cexde xx +=∫22 2

2. cedeewww +−=∫ cossin

3. ∫∫∫∫∫ +==⇒+=== cxdtdt

tdxdxcxdxxdxdx

)(1 0

19.3.2 Elementary integration rules

1. ∫∫ = dxxfdxxf )()( αα - constant factor out

linearity of integration

2. ∫∫∫ +=+ dxxgdxxfdxxgxf )()()]()([ - sum rule

These rules can be proven using the definition of integral and rules for limits.

Examples:

1. Integrate ∫ dxx sin2

Solution Integration variable? x Integrand? xsin2 It is not a table function of x The last operation in the integrand? Multiplication Is one of the factors a constant? Yes, 2 What rule to use? The “constant factor out” rule

Constant factor α = 2, variable factor xxf sin)( =

Substituting α and )(xf into the “constant factor out” rule,

cxdxxdxx +−== ∫∫ cos2 sin2 sin2

=⇒ ln xdx

d

= x

60

2. Integrate ∫ + dxxx )cos(sin

Solution Integration variable? x Integrand? xx cossin + It is not a table function of x The last operation in the integrand? Addition What rule to use? The sum rule

First term xxf sin)( = , second term xxg cos)( =

Substituting )(xf and )(xg into the sum rule,

cxxcxcxdxxdxxdxxx +−=+++−=+=+ ∫ ∫∫ cossinsincos cos sin)cos(sin 21

19.3.3 Integration Decision Tree

Similarly to the Decision Tree for Limits and Decision Tree for Differentiation, the Decision Tree for Integration (figure 19.1) allows you to decide at each step which Integration Rule or Integration Table entry to use when finding an indefinite integral. The decision depends on the last operation in the integrand. Do not forget to start at the top of the Tree and then follow the arrows that are associated with the correct answers (if any). After you have gone through the Decision Tree once check whether the operation of integration has been carried out. If not, hopefully, now you have simpler functions to integrate. In order to do that go over the Decision Tree again, substituting new integrands for f(x).

Figure 19.1. Decision Tree for Integration (incomplete).

The answer is

in the table

sum rule constant factor out

change of variable…


∫ ?)( dxxf

What is the integration variable? (appears after d) What is the integrand? (appears between ∫ & d)

Is f( ) a table function of integration variable x?

What is the last operation in f(x)?

Maybe, tricks

x

÷÷÷÷

Maybe, tricks

No

Yes

No

+ -

Yes

Other

61

19.3.4 Integration method (rule) of change of variable (substitution)

There are many other integration rules apart from the elementary ones considered in Section 19.3.1. We will study them later. They all follow from the rules for differentiation but are more complicated. First we introduce the simplest of them. It is called the change of variable method. Examples:

1. cxcudu

udxx ++=+=∫=∫ + 7.3)13(1.11

1

7.3

7.3

3

1

3

7.27.2)13(

Note: dx

duis not an ordinary fraction but a limit of a fraction. Since the limit of a fraction

is a fraction of limits it can be proven that we can formally “multiply” dx

du by dx to

get du. Question: What is du equal to? Answer: Question: What is dx equal to? Answer:

2. 129.0)4cos9(cos2

1cos

2

1sin

3

2

23

2

2 ≈−−=−=∫ tdttt

222 du

t dt, tdtdt

du,tu ===

ctcuududttt +−=+−== ∫∫22 cos

2

1cos

2

1sin

2

1sin

Integration variable? x Integrand? (3x + 1)

2.7

The integrand is not a table function of x. The last operation in the integrand? Power (hence the elementary rules do not apply). Introduce a new variable that lumps together the first few operations in the integrand

u = 3x + 1 dx

d, 3=

dx

du dx , 3÷ ,

3

dudx =

62

3. 0766.0)1cos2(cos3

1cossin 33

2

1

2 ≈−−=∫ tdtt

4. ceceduedxx

e xuux

+=+==∫∫tan

2

tan

cos

5. cxxu

dudx

xx

x+++==

++

+∫∫ 2ln

2

13 3

3

2

6. cxu

du

x

dxdx

x+−==

−=

− ∫∫∫ 75ln5

4

5

4

754

75

4

7. ( ) ctdtt

t++=

+∫ 1ln

2

1

1

22

8. cedt

e

et

t

t

+

+=

+

∫ 1ln2

1

2

2

2



• Revise Lecture 4 and Solutions to Solutions to Exercises in Lecture 4 (particularly, the operation of composition)



ct

cu

duutdtt +−=+−=−= ∫∫ 3

cos

3cossin

3322

u = tan x, dxx

duxdx

du

22 cos

1 ,

cos

1== ,

u = x3

+ x + 2, du = (3x2

+1) dx

u = 5 x − 7, du = 5 dx

u = cos t, du = − sin t dt

63



• Attempt the following exercises: Q1. Find

a) ∫ duu

1

b) ∫ dxu

1

c) )(1

tudtu

++∫

d) ∫ tdet

sinsin

e) ∫ttn

deesinsin

Q2. Find

a) ∫ + duuu

)cos1

(

b) ∫ + duuu )(2

c) ∫ ++ duuuu )2(32

d) ∫ ++ duvvv )532(52

Q3. Find

a) ∫ dttetsin

cos

b) ∫ −dt

t3

4

c) ∫− ++

+1

1

52 532

62dv

uvv

v

Additional Exercises Q4. Use the change of variable (substitution) to find

a) ∫ + dxx7

)14(

b) ∫ + .)1sin( 32 dttt Hint: let 13 += tu

c) ∫ − dxx )13sin(

d) ∫−

dxex 32

e) ∫ + dxxx42

)72(

64

f) ∫ tdtt 4cos4sin2

Q5 Evaluate

a) ∫−

4

32 3

dt

t

t

b) .ln

2

0

dzz

z∫ Hint: let zu ln=

Q6. Find

a) ∫++

+dx

xx

x

52

233

2

b) ∫3

2

32 sin dttt

c) ∫ −du

u 1

1

d) ∫ −du

u1

1

65

Lecture 20. INTEGRAL CALCULUS: Advanced Integration Methods

In Lecture 19 we have introduced the table of indefinite integrals of elementary functions and two simple linearity rules: one for integrating a sum of two functions and one for integrating a product with one of the factors constant. We now introduce additional rules (or methods) for integrating products and quotients of elementary functions. 20.1 Integration of products of trigonometric functions

1. ∫ =dtt2cos

=+=+=+=+

∫∫ ∫ ∫∫ )2cos(2

1)2cos(

2

1)2cos1(

2

1

2

2cos1tdtttdtdtdttdt

t

Question: Is cos2

t a product? Answer: No, but it can easily be re-written as a product, cos

2 t=cos t × cos t.

The square of a cosine is simply related to the cosine of a double angle

ttt22

sincos2cos −= , see the Trigonometry Summary.

We can eliminate sin2

t above using Pythagoras’ Theorem in terms of angles. Question: What is Pythagoras’ Theorem in terms of angles?

Answer: 1cossin22 =+ tt

tt22 cos1sin −=⇒

Substituting sin2

t into the double angle identity above gives us

1cos2cos1cos)cos1(cossincos2cos 2222222 −=+−=−−=−= tttttttt .

We can now use the Decision Tree for Solving Simple Equations to express

in cos t in terms of cos 2t, a simpler function of t,

2

2cos1cos2 t

t+

= .

The latter expression involves simple operations (on cos 2t) of addition and division by a constant, so we can integrate using the linearity rules.

The last integral is not in the Integration Table for elementary functions introduced in Lecture 17, but it can be found in Integration Tables in other textbooks or on the internet. The integral can also be evaluated using the Integration Table for elementary functions and change of variable method presented in Lecture 17, with u = 2t. Whatever method is used,

∫ += ct

t2

2sin2cos .

Question: What is a derivative of t2cos ? Answer: t2sin2− Verbalise: Thus, when differentiating cos, a constant factor in the argument

becomes a factor in the result and when integrating cos, it appears in the denominator of the result.

66

+t(2

1 c

t+)

2

2sin = +t

2

1 c

t+

4

2sin

using Pythagoras’ Theorem

2. ∫∫∫∫ +−=−=−= ,4

2sin

2 cos1 )cos1( sin

222c

ttdttdtdttdtt where we used the result

obtained in Example 1.

3. ])sin()sin(

[2

1])cos()[cos(

2

1sinsin

nm

tnm

nm

tnmdttnmtnmntdtmt

+

+−

−

−=+−−= ∫∫ +c,

where we used a trig identity for a product of sinuses, which can be obtained by combining identities for cosine of a difference, cos(A − B), and cosine of a sum, cos (A + B). 20.2 Integration by parts (integration of products of different types of functions)

Examples:

1. ∫ ∫ ++=−= cxxxxdxxxxdxx cossinsinsincos

Note 1: Adding c at this stage would not change the final answer.

Integration by Parts formula: ∫ ∫−= vduuvudv

u - line: u = x, 1

dx

du= , du = dx

v – line: dv = cos dx ∫, v = ∫ cos x dx = sin x

Note: can check that putting +c here would not change the final result. Substitute u , du, v and dv into the Integration by Parts formula.

Optional The product rule for differentiation gives us a product rule for integration (traditionally called integration by parts)

dx

duv

dx

dvu

dx

vud+=

⋅ )( dx

Question: What is dx? Answer: Multiplication by dx is purely formal, its meaning can be clarified by using properties of limits ⇒ d (u v) = u dv + v du

⇒ u dv = d (u v) − v du ∫ ⇒

67

Note 1: we reduced our task to integrating cos x – that is, to integrating a part of (the old

term for a factor in) the original integrand. But note that ∫ ∫∫ ⋅≠ xdxcosxdxxdxcosx

(that is, an integral of a product is not a product of integrals!) Note 2: The only challenge when using integration by parts is deciding which part of the integrand should be denoted by u. The rest is automatic.

2. cxxxdxxxx

xdxx +−=−= ∫∫ 164

ln

444

lnln

44443

Note: xln is a good choice for u, because u should be differentiated and derivative of

xln is x

1, which leads to simplification when multiplying by the integral of power 3

x (which

in itself is a power of x).

3. cxxxdx

xxxxdx +−=−= ∫∫ ln4

lnln

4. =+−= ∫∫ xdxxxxxdxx coscossin22

u - line: x

dxdu

xdx

duxu === ,

1 ,ln

v – line: dv = dx ∫, v = x

u - line: x

dxdu

xdx

duxu === ,

1 ,ln

v – line: dv = x3 dx ∫, v =

4

4x

u - line: x

dxdu

xdx

duxu === ,

1 ,ln

v – line: dv = dx ∫, v = x

Note: 2x is a good choice for u, because u should be differentiated and

derivative of 2x is x2 , a lower power, leading to a simpler integrand. Let us

continue by evaluating ∫ xdxx cos by parts:

68

∫ ++−=+++−=−+−= cxxxxcxxxxxdxxxxxx sin)1(coscossincos sinsincos 222

5. ∫∫ =−= dxxe

xdxxx

3x3

23x2 e3

2

3e

cxxe

cexee

xdxe

xe

xx

xxxxx

++−=++−=+− ∫ )269(2727

2

9

2

3e

3

2

33

2

3

23

333

23x33

2

6. ∫∫ =+−= dxxxdxx e cos3)cos(e e sin 3x3x3x

∫−+− ) sin3(sin3cos 333 dxxexexe xxx

⇒ ∫∫ −−= xdxsine9)xcosxsin3(exdxsine x3x3x3

⇒ I = x3e (3 sin x – cos x ) – 9 I

⇒ I = xe

3

10

1(3 sin x – cos x ) + c (since indefinite integrals are defined up to a

constant term)

20.3 Partial fractions The method of partial fraction is an algebraic trick that allows us to re-write a rational function as a sum of the simplest possible rational fractions. It is inverse (opposite) to adding fractions using the common denominator method.

Consider ))((

1122 axaxax +−

=−

= ax

A

− +

ax

B

+

⇒ A (x + a) + B (x – a) = 1

To solve this equation for A and B use one of the two methods: I. Remove the brackets A x + Aa + B x – Ba = 1

(x + a) (x − a)

u - line: u = x, 1=dx

du, du = dx

v – line: dv = cos x dx ∫ , v = ∫ dxx cos = sin x

Substitute u, du, v and dv into the Integration by Parts formula to obtain

u

u

u

I I

u

69

Collect the like terms (A + B)x + Aa – Ba = 1 for all x Equate the like coefficients

⇒ A + B = 0

Aa – Ba = 1

⇒ A = –B ⇒ –Ba – Ba = 1 ⇒ B = –a2

1, A =

a2

1

OR II. Use the killer instinct (!)

x = –a: B (–2a) = 1 ⇒ B = –a2

1 (chose a value of x to “kill” the first term)

x = a: 2aA = 1 ⇒ A = a2

1 (chose a value of x to “kill” the second term)

Whatever method you use, you can see that

)11

(2

11

22 axaxaax +−=

−

Check: 222222

12

2

1)(

2

1)

11(

2

1

axax

a

aax

axax

aaxaxa −=

−=

−

−−+=

+−

−

Representing a rational function as a sum of partial fraction is useful for Integration of rational functions. 20.4 Integration of rational functions

Example: ∫ =−−

dxxx )3)(2(

1

)( )( axax −+

)3x)(2x(

1

−− =

2−x

A +

3−x

B =

)3)(2(

)2()3(

−−

−+−

xx

xBxA

⇒ (A + B) x – (3A + 2B) = 1

Equating the like coefficients, A + B = 0 ⇒ A = –B

3A + 2B = − 1 ⇒ –3B + 2B = –1 ⇒ –B = –1 ⇒ B = 1 ⇒ A = –1

Check: )3)(2(

1

)3)(2(

)2()3(

3

1

2

1

−−=

−−

−+−−=

−+

−−

xxxx

xx

xx�

(x -3) (x - 2)

)3( )3( −− xx

70

∫ ∫ +−

−=+−+−−=

−+

−−= c

x

xcxx

x

dx

x

dx

2

3ln3ln2ln

32

20.5 Decision Tree for Integration We can now introduce the full Decision Tree for Finding Indefinite Integrals – see figure 20.1:

Figure 20.1. Decision Tree for Integration. 20.6 Instructions for self-study

• Revise ALGEBRA Summary (addition of fractions)

• Revise Summaries on FUNCTIONS and TRIGONOMETRY

• Revise Lectures 14 - 15 (limits and differentiation)

Maybe, tricks

Yes No

Yes Yes No No

No Yes

Yes No

+ - Other

x

÷

( )∫ dxxf ?

What is the integration variable? Look after d

What is the function to integrate? Look between ∫ & dx Maybe, tricks

Is f ( ) a table function of the integration variable?

Is one of the factors const

1≠ ?

What is the last operation in f (x)? Answer

Separate terms

Rational function? Constant factor out

Trig. function? Proper?

Trig. identities or change of variable

Change of variable or integration by parts

Partial fractions or change of variable

Polynomial plus partial fractions or change of variable

71

• Revise Lectures 18 - 19 (integration) and study Solutions to Exercises in Lecture 18 using the STUDY SKILLS Appendix


• Do the following exercises: Q1. Find

a) ∫π2

0

2sin tdt

b) ∫π2

0

2cos tdt

c) ∫ xdxx cossin

d) ∫ xdxx 2coscos

Q2. Find

a) ∫ xdxx sin

b) ∫ xdxx ln2

c) ∫ dxex x32

d) xdxe x cos2

∫

Q3. Find

a) ∫ −−dx

xx )3)(1(

1

b) ∫ +−

+dx

xx

x

23

12

c) ∫ −−

−dx

xx

x

82

12

d) ∫ +−

−dx

xx

x

82

12

72

Lecture 21. INTEGRAL CALCULUS: Applications of Integration 21.1 Mean value of a function Using The Mean Value Theorem (Lecture 18), the mean value of a function on the interval between a and b is

M ≡ f (c) = ∫−

b

a

dxxfab

)(1

(mean height)

⇒Area of dashed rectangle = Area under the curve

M(b - a) = ∫b

a

dxxf )(

Optional RMS value of a function (the square root of the mean value of the squares of y) Example: Find RMS of y = x

2 + 3, 31 ≤≤ x

=

21

3

1

242

13

1

2 )96(2

1

13

1

++=

− ∫∫ dxxxdxy

=

21

3535

2/1

21

3

1

35

93

1

5

127

3

3

5

3

2

19

352

1

++−

++=

++ x

xx≈ 2

1

2/1]74[

2

1637 ≈≈

a b x

y

M

Example: Find mean of y = 3x2 +4x + 1, 21 ≤≤− x

Solution a = − 1, b = 2

M = )23(3

1)143(

)1(2

1)(

1 32

1

2 xxxdxxxdxxfab

b

a

++=++−−

=− ∫∫

−

= 6

73

21.2 Electrical and mechanicsl systems Electrical systems If charge q varies linearly with time t, then current i is constant with time

t

qqi 0−

=

If charge q is non-linear in time t, then instantaneous current i is

dt

dqi = ⇒ q(t) = )0()(

0

qdi

t

+∫ ττ

Mechanical systems If displacement x varies linearly with time t, then speed v is constant with time

t

xxv 0−

=

If displacement x is non- linear in time t, then instantaneous speed v is

dt

dxv = ⇒ x(t) = )0(

0

xv( τ(τ)t

+∫

Example: Consider a capacitor of capacitance C (in Faradays). The voltage drop on the capacitor is

C

qdi

CC

qe

t

c0

0

)(1 (t)

(t) +== ∫ ττ . Below we often assume q0 = 0.

Let i = i0 cos ωt, T

πω

2 = - circular frequency, T - period

⇒ tC

id

C

ie

t

c ωω

τωτ sincos 0

0

0 == ∫ )2

cos()2

cos( 00 πω

ωω

π

ω−=−= t

C

it

C

i

⇒ a capacitor effects a 2

π shift between current and voltage

π

ωt

eC

i

i0

ωt

ωC

i0

q0

t

q

q0

t

q

x0

t

x

x0

t

x

q0

q

t

74

21.3 Rotational systems 21.3.1 Volumes of Solids of Revolution

If the plane figure bounded by the curve y = f(x), the x-axis and the straight lines ax = and

bx = rotates through a complete revolution about the x-axis, it generates a solid. Let V be its volume.

To find V we represent the area under the curve as a sum of areas of thin strips of width x∆ (see figure above). The volume is approximately a sum of volumes of thin cylinders generated by each of these thin strips, Question: What is the volume of a cylinder? Answer: Question: In our case, what is the radius r at the base of each thin cylinder? Answer: Question: What is the height h of each thin cylinder? Answer:

⇒ ∑ ∑ ∫ →∆ =∆≈n n

b

a

dxyyV22 xV ππ as 0→∆x

Example: Find the volume generated when the plane figure bounded by xy 2cos5= , the

x-axis and the straight lines 0=x and 4

π=x rotates through a complete revolution about

the x-axis.

301.34

)4

4sin(

45.12

04

)4sin(5.12

2

)4cos(125)2(cos25)]2cos(5[

24

4/

0

4/

0

24/

0

22

≈≈

⋅

+=

+

=+

==== ∫∫∫∫

π

ππ

ππ

ππππ

π

πππ

xx

dxx

dxxdxxdxyV

b

a

y = f(x)

x

y = f(x)

a x∆ b x

75

21.3.2 The moment of inertia The moment of inertia I of a point of mass m rotating around a center O is I = mr

2, where r is the distance between the point and the centre. The moment of inertia of a lamina around an axis is approximately the sum of the moments

of inertia ∆I = r2∆m of elementary points of mass ∆m lying at the distance r from the axis of rotation,

∑ ∑ ∫ →∆ =∆≈n n surface

n drrI mmI 22 as .0→∆x

Example:

Let a square lamina rotate around the y-axis (see figure below). All points in the thin shaded area are approximately the same distance x from the axis.

If mass density (mass per unit area), ,ρ of the lamina is constant (does not change

with position), then the mass of the shaded strip is ∆m = xa∆ρ2 (density times area),

and the moment of inertia of the lamina is

322

322 x

aadxxdmrI

a

asurface

ρρ === ∫∫−

= 33

4))((

3

2 2433 Maa

aaa

==−−ρρ

,

where the lamina mass is M = 24 aρ .

Question: Why is the lamina mass 24 aρ ?

Answer:

O m

a

− a

-a a x

∆x

y

76


• Revise Lectures 14 - 15 (differentiation)

• Revise Lectures 19 and study Solutions to Exercises in Lecture 19 using the STUDY SKILLS Appendix



• Do the following exercises:

Q1. Given titi sin)( 0 ω= find eC and eL - provided ω0

0

iq −= , that is

C

i

C

qeC

ω00)0( −== .

Q2. Find the volume generated when the plane figure bounded by xy 4= , the x-axis and

and the straight lines 0=x and 1=x . Q3. a) The moment of inertia about a diameter of a sphere of radius 1 m and mass 1 kg is found by calculating the integral

∫− −1

1

22 )1(8

3dxx .

Show that the moment of inertia of the sphere is 5

2kg m2.

b) Calculate the moment of inertia of a uniform thin rod of mass M and length l about a perpendicular axis of rotation at its end.

77

Lecture 22. Ordinary Differential Equations 22.1 Basic concepts Consider the equation

)()(

xaydx

xdy= (22.1)

Eq. (22.1) is an example of an ordinary differential equation. It is called an equation, because it is a mathematical statement that contains the = sign and that can be true or false depending on which function you substitute for the unknown y(x) (compare to the

case of an algebraic equation 0)( =xPn : an algebraic equation is a mathematical

statement that contains the = sign and can be true or false depending on which value you substitute for the unknown x). Terminology: y(x) is called an unknown (function) or dependent variable; x – an independent variable; a - a (constant) coefficient. Eq. (22.1) is a differential equation, because it contains a derivative of the unknown y. It is called ordinary, because the unknown is a function of one variable (the differential equations with unknowns being functions of several variables are called partial, because they contain partial derivatives). Application: Eq. (22.1) describes many systems in which at any given time, the rate of signal change (that is, derivative of the signal) is proportional to the value of this signal. To solve a differential equation means to find all functions which turn it into a true statement (compare to the case of an algebraic equation: to solve an algebraic equation means to find all values of the unknown variable which turn it into a true statement). Examples:

1. In the following equations, state which variable is dependent and which variable is independent:

a) tdt

dy

dt

yd=+

2

2

. Answer:

b) 0vvdt

dvRC =+ Answer:

c) .0)( =+ ytcdt

dyy Answer:

2. Solve xedx

xdy=

)(

The above equation is simple – only one term contains the unknown function y(x). Therefore, it can be solved using the Decision Tree for Solving Equations with Only One Term Containing the Unknown. In general, to solve a differential equation it is important to classify it first, because different types of differential equations can be solved using different methods (tricks). The differential equations are classified according to their order, to whether they are linear or non-linear and to whether their coefficients are constant or variable.

78

22.2 Order of a differential equation The order of a differential equation is the order of its highest derivative.

Example: Establish the order of the above differential equations. Answers:

22.3 Linearity or non-linearity of a differential equation A differential equation is called linear if each of its terms either contains no unknown function or is a product of the unknown function or one of its derivatives and a known coefficient (constant or variable).

Example: Establish which of the above differential equations is linear and which non-linear. Answers:

22.4 Differential equations with constant coefficients A differential equation has constant coefficients if in each term the coefficients (factors) of the dependent variable and its derivatives are constant (with respect to the independent variable).

Example: Establish which of the above differential equations a) – c) has constant coefficients. Answer:

22.5 Homogeneous and inhomogeneous ODEs A differential equation is called homogeneous if each of its terms contains the unknown or its derivatives. Otherwise it is called inhomogeneous.

Example Establish which of the differential equations a) – c) is homogeneous. Answer:

22.6 The first order linear homogeneous equation with constant coefficients Eq. (22.1) is a general form of the first order linear homogeneous equation with constant coefficients. We can check by substitution that y = e

ax is its solution. We can check by substitution that y = ce

ax with any arbitrary constant factor c is also its solution.

79

⇒ Any first order linear homogeneous equation with constant coefficients has infinitely many solutions involving one arbitrary constant. 22.7 The initial value problems In reality, the initial value of an unknown signal is often known. If given an initial condition,

say 0)0( yy = , the solution becomes unique.

The problem

=

=

0)0(

'

yy

ayy

is an example of an Initial Value Problem. 22.8 Balance equations in chemical engineering

For every element of a chemical system there will be a material balance equation for each chemical species present:

∑∑ −= )rates outflow()rates inflow()quantity(

dt

d

In this equation quantity must be an extensive quantity, e.g. mass (kg), energy (Joules) etc. Do not try to write material balance equations for temperature or concentration. Flow rates must be measured, respectively, in kg/s, Watts, etc.

Example:

Consider a tank containing V litres of a solution consisting of x0 (kg) of salt dissolved In water. At the initial moment of time t = 0, let us start pumping pure water into the tank at the rate of r (litres/s) and let x(t) (kg) be the mass of salt in the tank at any moment t. Let us keep the mixture uniform by stirring. Let the outflow rate of mixture be the same as the inflow rate of water, that is, r. Then the volume V of mixture is kept constant throughout. It is easy to check that under these conditions the outflow rate of salt is

V

txr

)((kg/s)

and the amount of salt x(t) satisfies the ordinary differential equation

V

x(t)rtx' −=)(

The associated Initial Value Problem is

=

−=

00

)(

x)x(

x(t)V

rtx'

80

22.9 Ordinary differential equations with complex coefficients Consider an equation jyy =' (22.2)

Using Section 22.6, the function jteyy 0= is its solution.

Optional

We can also prove this without using Section 22.6 as follows:

Substituting y = y1 + jy2 into the right-hand side of Eq. (22.2) we obtain:

21 yjyy ′+′=′

Substituting y = y1 + jy2 into the right-hand side of Eq. (22.2) we obtain:

jy = j y1 - y2

This means that Eq. (22.2) can be written as

2121 yjyyjy −=′+′ (22.3)

At each moment t, Eq. (22.3) has one complex number on the left-hand side and another, in the right-hand side. This means that their real parts are equal and so are their imaginary parts,

12

21

yy

yy

=′

−=′

Let .cossin 02102 tyyytyy =′=⇒= This means that

jteytjytyjyyy 00021 sincos =+=+= ,

where we used the Euler’s formula.

81

Note: When differentiating or integrating, complex coefficients must be treated in the same manner as real. 22.10 Applications Eq. (22.1) describes many systems in which at any given time, the rate of change of the variable (that is, derivative of the variable) is proportional to the value of this variable. Ordinary differential equations and associated Initial Value Problems are common in science and engineering, because scientific and engineering systems often effect transformations of signals, which involve their rates of change, that is their derivatives. In many of your science and engineering courses you will be using linear ordinary differential equations of second order, with constant coefficients. 22.11 Instructions for self-study

• Revise Decision Tree for Solving Simple Equations

• Revise Summaries on COMPLEX NUMBERS and DIFFERENTIATION




• Do the following exercises: Q1. Verify that

a) y = 3 sin2x is a solution of 04y = +y ′′

b) xe2 satisfies equation 04y = +y ′′

c) xAe satisfies equation 02 +y=y-y ′′′

d) xxBxeAe + satisfies equation 02 +y=y-y ′′′

e) Classify all the above equations. Q2. Consider a circuit, comprising a capacitor of capacitance C and a resistor R placed in series: When a constant voltage source, V, is applied it can be shown that the current, I, through the circuit satisfies the ordinary differential equation.

01

=+ ICdt

dIR

Given the initial current values I(0) = VR solve the corresponding initial value problem, that is find the current I(t) through this circuit.

V - +

R C

82

Q3. Consider a tank containing 5 litres of a solution consisting of 0.5 (kg) of salt dissolved in water. Let pure water be pumped into the tank at the rate of 1 (litres/s) and the mixture, which is kept uniform by stirring, be pumped out at the same rate. Find x(t), the amount of salt in the tank at any moment t. Q4. Solve the equations

a) jyy 2=′

b) xy sin2=′′

83

IV. SUMMARIES Algebra Summary

OPERATIONS TYPES OF VARIABLES Addition (direct operation) Addition of whole numbers gives whole number 1. a + b = b + a

Terminology: a and b are called terms a + b is called sum 2. (a + b) + c = a + (b + c)

Subtraction (inverse operation) Def: a – b = x: x + b = a Note: a + b – b = a (subtraction undoes addition) a – b + b = a (addition undoes subtraction) 3. a + 0 = a

4. for each a there exists one additive inverse –a:

a + (–a) = 0

Rules (follow from Laws): + (b + c) = + b + c

+ a + b = a + b

– (– a) = a

– (a) = –a

Whole numbers are 1, 2, 3, ... 1 2 3 x

introduces 0 and negative numbers: a – a = 0

if b > a a – b = – (b – a)

Natural numbers are 0, 1, 2, ... 0 1 2 3 x Integers are ..., –2, –1, 0, 1, 2, ...

–2 –1 0 1 x

84

Multiplication (direct operation) For whole numbers n

a⋅n = a+...............+a

n terms

Notation: ab = a⋅b = a×b

2b = 2⋅b = 2×b

23 ≠ 2⋅3, 23 = 2⋅10+3

2

12 ≠

2

12 ⋅ ,

2

12 =

2

12 +

2

32 =

2

32 ⋅

1. a⋅b = a⋅b

Terminology: a and b are called factors ab - product

2. (a⋅b)⋅c = a⋅(b⋅c)

Conventions: abc = (ab)c

a(–bc) = –abc 3. a(b+c) = ab+ac

Removing brackets Factoring

4. a⋅0 = 0

5. a⋅1 = a

Rules (follow from Laws): (a + b)(c + d) = ac + ad + bc + bd (SMILE RULE)

(–1) ⋅ n = –n

(–1) ⋅ (–1) =1

Division (inverse operation) Def: a/b = x: xb = a Terminology: a – numerator b – denominator a/b – fraction (ratio)

- proper fraction if |a|<|b|, a, b integers Note: ab/b = a (division undoes multiplication) (a/b)b = a (multiplication undoes division)

6. For each a ≠ 0 there exists one

multiplicative inverse 1/a: a⋅1/a = 1

introduces rational numbers

Def: Rationals are all numbersn

m,

where m and n ≠ 0 are integers (division by zero is not defined)

4

1

3

1

2

1

–1 0 1 2 x

85

Rules:

b

a⋅n =

b

an

n

ba

= bn

a=

b

na

bn

an =

nb

na

⋅

⋅ =

b

a CANCELLATION

m

n

1

= n

m FLIP RULE

b

a+

b

c=

b

ca + Note:

b

ca += (a+c)/b

b

a+

d

c=

bd

ad+

db

cb=

bd

cbad + COMMON

DENOMINATOR RULE

d b

\

86

n-th power bn (direct operation)

For whole numbers n

bn = b⋅b⋅b⋅…⋅b

n factors

Rules:

am⋅ an

= am+n

(product of powers with the same base is a power with indices added)

an⋅ bn = (ab)

n

(product of powers is power of product)

n

m

a

a= a

m-n

(ratio of powers is power of ratios)

m

m

m

b

a

b

a

=

(ratio of powers with the same base – subtract indices)

a0

= 1

a-n

= na

1

(am)n

= amn [Convention: )( nn mm

aa = ]

n-th root (inverse to taking to power n)

Def: bxbn == n x:

Note: bbn n =

(taking n-th root undoes taking n-th power)

bb nn =)(

(taking n-th power undoes taking n-th root)

Therefore, can use notation nn bb =1

(Indeed, bbbbb nn

nnn n ====⋅

1

11

)( )

0 1 x

introduces irrational (not rational)

numbers 2 , 3 , 5 , 3 2 , 3 3 …

Real numbers are all rationals and all irrationals combined. Corresponding points cover the whole number line (called real line for this reason)

87

Logarithm base b (inverse to taking b to power)

Def: abnan

b == :log

Note: logb b

n = n (check using definition: bn

= bn)

(taking logb undoes taking b to power)

nbn

b =log

(taking b to power undoes logb) Rules (follow from Rules for Indices): logb xy = logb x + logb y

(log of a product is sum of logs)

logb y

x = logb x – logb y

(log of a ratio is difference of logs) logb1 = 0 (log of 1 is 0) logb b = 1

logba

1 = –logb a

logb xn = n logb x

(log of a power is power times log)

logb a = b

a

c

c

log

log

(changing base) General remarks 1. a – b = a + (–b) a difference can be re-written as a sum

2. b

a = a ⋅

b

1 = ab

-1 a ratio can be re-written as a product

3. nn bb1

= a root can be re-written as a power

4. All laws and rules of addition, multiplication and taking to integer power

operations apply to real numbers.

5. Operations of addition, subtraction, multiplication, division (by non-zero) and taking to integer power when applied to real numbers produce real numbers. Other algebraic operations applied to real numbers do not necessarily produce real numbers.

introduces irrational (not rational) numbers, log10 2, log10 3, etc. Roots and logs also introduce complex (not real) numbers,

)1(log,1 10 −− , etc.

88

Functions Summary

Variables are denoted mostly by x, y, z, p, ….. w. A variable can take any value from a set of allowed numbers. Functions are denoted mostly by f, g and h or f ( ), g ( ) and h( ) (no multiplication sign is intended!). In mathematics the word function has three meanings:

To specify a function we need to specify a rule - (series of) operation(s) and domain D (an allowed set of values of the independent variable). To each x∈D, f (x) assigns one and only one value y∈R (range, the set of all possible values of the dependent variable).

. Inverse functions f

-1 (x): f ○ f

-1 (x) = f

-1 ○ f (x) = x (the function and inverse function undo each other)

The inverse function does not always exist!

1. f( ) - an operation or a chain of operations on an independent variable;

2. f(x) - a dependent variable, that is the variable obtained when f( ) acts on an independent variable x;

3. Set of pairs {(x, f(x): f( ) assigns one value f(x) to every allowed value of x}.

A diagrammatical representation of a function

f D R

symbol of inverse function, not a reciprocal

A Diagrammatic Representation of Inverse Function

f -1

(y)

f(x) R

y D

x

y = f(x)

x = f -1

(y)

89

Order of Operations Summary When evaluating a mathematical expression it is important to know the order in which

the operations must be performed. By convention, the Order of Operations is as follows:

First, expression in Brackets must be evaluated. If there are several sets of brackets, e.g.

{[( )]}, expressions inside the inner brackets must be evaluated first. The rule applies not

only to brackets explicitly present, but also to brackets, which are implied. Everything

raised and everything lowered is considered as bracketed, and some authors do not

bracket arguments of elementary functions, such as exp, log, sin, cos, tan etc. In other

words, ex should be understood as exp (x), sin x as sin (x) etc.

Other operations must be performed in the order of decreasing complexity, which is

oiB - operations in Brackets

F - Functions f ( )

P- Powers (including inverse operations of roots and logs)

M - Multiplication (including inverse operation of division)

A - Addition (including inverse operation of subtraction)

That is, the more complicated operations take precedence. For simplicity, we refer to this

convention by the abbreviation oiBFPMA.

.

Order of operations (OOO)

1. Make implicit (invisible) brackets visible (everything raised and everything lowered is considered to be bracketed and so are function arguments)

2. Perform operations in brackets {[()]} first (inside out) OOO oiB f( ) P M A

(including ( )( ) × +

implicit ) roots ÷ - logs

90

Quadratics Summary A quadratic expression is a general polynomial of degree 2 traditionally written as

cbxax ++2 where a is the constant factor in the quadratic term (that is, the term containing the independent variable squared): b is a constant factor in the linear term (that is, the term containing the independent variable) and c is the free term (that is, the term containing no independent variable). A quadratic equation is the polynomial equation

02 =++ cbxax Its two roots (solutions) can be found using the standard formula

a

acbbx

2

42

2,1−±−

=

Once the roots are found the quadratic expression can be factorised as follows:

))(( 212 xxxxacbxax −−=++

91

Trigonometry Summary Conversion between degrees and radians

An angle described by a segment with a fixed end after a full rotation is said to be 360° or

2π (radians)

⇒ 2π (rad) = 360 0

⇒ 1 (rad) ≈ 57 0

Above, x is the number of radians (angles ≈ 57 0 ) in a given angle (cf: x (m) = x⋅ 1m, so that x is the number of units of length (segments 1m long) in a given segment)

(Note: cf. means compare)

⇒ x (rad) = x)(

1800

radπ= 0y , 0y =

0180

)(rady

π= x (rad)

Usually, if the angle is given in radians the units are not mentioned (since the radian is a dimensionless unit). Frequently used angles

180 0 0 0 (360 0 )

π 2π

30 0 = (180

30π=)

6

π

60 0 = (180

60π=)

3

π

120 0 = (180

120π=)

3

2π

45 0 = 4

π

The radian is a dimensionless unit of angle.

90 0

2

π

270 0

2

3π

92

Right Angle Triangles and Trigonometric Ratios

Trigonometric ratios sin, cos and tan are defined for acute angles (that is, angles less than 900) as follows:

b

aa

c

b

c

a

==

==

==

β

βα

βα

cottan

sincos

cossin

α + β = 900 and α and β are called complementary angles

Frequently used trigonometric ratios

2

1

6sin =

π

2

2

4sin =

π

cos 6

π =

2

3

2

2

4cos =

π

2

2

4cos =

π

Trigonometric identities

sin

2 x + cos

2 x = 1

sin ( − x) = − sin x

cos ( − x) = cos x

tan ( − x) = − tan x yxyxyx sinsincoscos)cos( m=±

yxyxyx sincoscossin)sin( ±=±

xxx22 sincos2cos −=

xxx cossin22sin =

)sin(1

sincos22

α++

=+ x

BA

xBxA , where B

A=αtan

a

c

β

α b

93

Complex Numbers The Cartesian representation of a complex number

The exponential representation of a complex number

z= x+jy, where x,y – real, j = 1−

y (z) z1 (x1, y1)

0 x1 x

real part, Re z imaginary part, Im z (on the Argand diagram)

x-coordinate y-coordinate (on the ordinary plane)

r1

θ1

z1

y (z)

0 x

z = r e jθ , where r ≥ 0, θ - angle (in degrees or radians)

magnitude, |z| argument, arg(z) (on the Argand diagram)

polar radius r polar angle θ (on the ordinary plane) (distance to the origin)

94

Decision Tree For Solving Simple Equations

No

No

Yes

Yes

Yes

Is the answer a proper fraction?

What is the highest common divisor for numerator and

denominator?

What is the integer part of the fraction?

What is the remainder?

Answer

No

If needs be, use various rules to ensure that the unknown appears only in one term in

the LHS (left-hand side)

What is the last operation on the unknown?

What is its inverse?

Apply this inverse operation to both sides of the equation

Is the unknown the subject of the equation?

Is the answer in its simplest form?

95

Sketching Graphs by Simple Transformations 1. Drop all constant factors and terms. 2. Bring the constants back one by one in the Order of Operations (not necessary but

advisable) and at each Step use the Decision Tree given below to decide which simple transformation is effected by each constant. Sketch the resulting graphs underneath one another.

DECISION TREE FOR SKETCHING y=f(x)+c, y=f(x+c), y=cf(x) AND y=f(cx)

Note 1: if c is a negative factor, write c = ( − 1)*│c│, so that c effects two simple transformations and not one. Note 2: if c is effects neither first operation nor last, the Decision Tree is not applicable. Try algebraic manipulations. Note 3: if the same operation is applied to x in all positions in the equation this operation still should be treated as the first operation, as in ln (x − 2)+sin (x − 2). Note 4: if y is given implicitly rather than explicitly, so that the equation looks like f(x,y)

= 0, then in order to see what transformation is effected by adding a constant c to y or multiplying it by a constant c, y should be made the subject of the functional equation. Therefore, similarly to transformations of x above, the corresponding transformation of y is

defined by the inverse operation, − c or c

1, respectively.

c > 0

c = -1

Is c > 0 or c= − 1 ?

Translation wrt y-axis by

c

Translation wrt x-axis by − c

Reflection wrt x-axis

Reflection wrt y-axis

Scaling wrt y-axis

by c

Scaling wrt x-axis

by c

1

Is +c the first operation or

last operation?

Is *( − 1)

the first operation or last operation?

Is *c the first operation or

last operation?

last

last

last

first

first

first

Is the constant c a factor or term?

Term

Factor

96

Finding a Limit of a Sequence To find a limit of a sequence (as ∞→k ) use the first principles (graphical representation) or TABLE RULES

DECISION TREE

Maybe, tricks

xk

lim xk

constant constant

k ∞

k

1

0

na

1

(a > 0)

1

qn (|q| < 1) 0

1. lim αxk = α lim xk – constant factor out

2. lim [xk±yk] = lim xk ± lim yk – sum rule (separate terms)

3. lim (xk yk) = lim xk lim yk – product rule

4. k

k

k

k

y

x

y

x

lim

limlim = – quotient rule

Answer

No Yes

Yes No

Maybe, tricks


What is the last operation in xk?

Is xk a table sequence?

Yes No

+ -

Other

separate terms

const factor

out

An indeterminate form?

(∞ − ∞, 0⋅∞, 0/0, ∞/∞, etc.)

if allowed reverse order of lim- and subsequent operation

Use a trick

xk ?

x

÷

97

Differentiation Summary To differentiate a function use the first principles

Def: x

xfxxf

x

y

dx

dy

xx ∆

−∆+=

∆

∆=

→∆→∆

)()(limlim

00

or represent roots and fractions as powers, make invisible brackets visible and use

TABLE RULES

f (x) )(xf

dx

d

constant 0

xn nx

n-1

ex

ex

ln x

x

1

sin x cos x

cos x − sin x

x

÷

Maybe, tricks Maybe, tricks

No Yes Yes No


Is y(x) composite?

What is the last operation in y(x)?

No

The answer is

in the table

Is y(x) a constant or another table function of x?

Product rule or

quotient rule

Constant factor

outside Chain rule Use tricks

Yes

Other ±

?)(xydx

d

What is the differentiation variable ? (it appears after d at the bottom)

What is the function to differentiate ? (it appears after d at the top)

Sum rule

f(x)

f(x+∆x)

y = f(x)

x x+∆x x

DECISION TREE

1. dx

xdfxf

dx

d )()]([ αα = – constant factor out

2. dx

xdg

dx

xdfxgxf

dx

d )()()]()([ +=+ – sum rule (separate terms)

3. )()(

)()(

)]()([ xfdx

xdgxg

dx

xdfxgxf

dx

d+= – product rule

4. )(

)()(

)()(

)(

)(2

xg

xfdx

xdgxg

dx

xdf

xg

xf

dx

d−

= – quotient rule

5. dx

xdg

xdg

xgdfxgf

dx

d )(

)(

))(())(( = – chain rule

(decompose, differentiate, multiply)

98

Integration Summary To find a definite integral of a function use the definition

Def: ∑ ∆==→∆

∫N

kk

b

a x

xxfdxxf00

)( lim)(

or represent roots and fractions as powers, make invisible brackets visible and use

TABLE RULES

DECISION TREE

)(xf ∫= dxxfxF )()(

0 C

ex e

x +c

cos x sin x +c

sin x − cos x +c

xn, n≠ − 1

cn

xn

++

+

1

1

0,1

≠xx

ln│x│+c

f(xk)

y = f(x)

a xk b x

Maybe, tricks

Yes No

Yes Yes No No

No Yes

Yes No

+ - Other

x

÷

∫ ?)( dxxy

What is the integration variable? Look after d

What is the function to integrate? Look between ∫ & dx Maybe, tricks

Is y( ) a table function of the integration variable?

Is one of the factors const

1≠ ?

What is the last operation in y(x)? Answer

Separate terms

Rational function? Constant factor out

Trig. function? Proper?

Trig. identities or change of variable

Change of variable or integration by parts

Partial fractions or change of variable

Polynomial plus partial fractions or change of variable

1. ∫∫ = dxxfdxxf )()( αα - constant factor out

2. ∫∫∫ +=+ dxxgdxxfdxxgxf )()()]()([ - sum rule

(separate terms)

99

V. GLOSSARY ABSTRACTION - a general concept formed by extracting common features from specific examples. ACUTE ANGLE – a positive angle which is smaller than 90

0. ALGEBRAIC OPERATION – OPERATION of addition, subtraction, multiplication, raising to power, extracting a root (surd) or taking a log. ALGORITHM – a sequence of solution steps. ARGUMENT – INDEPENDENT VARIABLE, INPUT. CANCELLATION (in a numerical fraction) – operation of dividing both numerator and denominator by a common divisor. CIRCLE – a LOCUS of points inside and on a CIRCUMFERENCE. CIRCUMFERENCE – a LOCUS of points at the same distance from a specific point (called a centre). COEFFICIENT – a CONSTANT FACTOR multiplying a VARIABLE, e.g. in the EXPRESSION 2ax, x is normally a VARIABLE and 2a is its COEFFICIENT. COMPOSITION – a combination of two or more functions (operations), with one acting on the output of another, e.g. composition of two functions f ( ) and g ( ) is f (g ( )). CONCEPT - a technical word or phrase. CONSTANT – a number or a mathematical quantity that can take a range of (numerical) values but is independent of the main CONTROL VARIABLE – ARGUMENT or UNKNOWN. If there is only one CONSTANT, the preferred choice for its symbol is a. The second choice is b, and the third, c, so e.g. in the EXPRESSION x + a, x is usually understood to represent a VARIABLE and a, a constant. If there are more CONSTANTS in the EXPRESSION, then one chooses, in the order of preference, d and e, and then upper case letters in the same order of preference. If more CONSTANTS are required, we make use of subscripts and superscripts and Greek letters. DIAGRAM - a general (abstract) visualisation tool, a pictorial representation of a general set or relationship. DIFFERENCE – a mathematical expression in which the last operation is subtraction. DIMENSIONAL QUANTITY – a quantity measured in arbitrary units chosen for their convenience, such as s, m, N, A, m/s, kg/m3, £. DOMAIN – a set of all allowed values of the ARGUMENT.

100

EQUATION – a mathematical statement involving UNKNOWNS and the = sign which can be true or false, e.g. 2x + 3y = 10 is true when x = y = 2 and not true when x = y = 1. EVALUATE – find the (numerical) value of a (mathematical) EXPRESSION. EXPLICIT – 1) clearly visible; 2) a subject of equation. EXPONENTIATION – a mathematical operation of raising to power. EXPRESSION (mathematical) – a combination of numbers, brackets, symbols for variables and symbols for mathematical operations, e.g. 2(a + b), 2ab. FACTOR – a (mathematical) EXPRESSION which multiplies another (mathematical) EXPRESSION, e.g. ab is a PRODUCT of two FACTORS, a and b. FINAL (SIMPLEST) FORM (of a numerical fraction) – no CANCELLATIONS are possible, and only PROPER FRACTIONS are involved. FORMULA – a mathematical statement involving VARIABLES and the = sign which and is always true. FREE TERM – a CONSTANT TERM. FUNCTION

1. f( ) - an operation or a chain of operations on an DEPENDENT VARIABLE.

2. f(x) - a DEPENDENT VARIABLE, that is the variable obtained when f( ) acts on an independent variable x.

3. Set of pairs {(x, f(x)): f( ) assigns one value f(x) to every allowed value of x}.

To give an example, ‘f (x) = 2x + 3, x - real' describes a FUNCTION whose SYMBOL is f or f ( ) (no multiplication is implied) and whose ARGUMENT is called x. Its DOMAIN is 'all reals'. Given any value of x, you can find the corresponding value of this FUNCTION by first multiplying the given value of x by 2 and then by adding 3 to the result. FUNCTION SYMBOL – usually a Latin letter, usually from the first part of the alphabet. If there is only one FUNCTION, the preferred choice for its symbol is f. The second choice is g, and the third, h. If there are more variables, then one chooses, in order of preference, u, v and w. If more FUNCTIONS are involved, we often make use of subscripts and superscripts, upper case and Greek letters. GENERALISATION - an act of introducing a general concept or rule by extracting common features from specific examples. GRAPH - a specific visualisation tool, a pictorial representation of a particular set or relationship. IDENTITY – the same as FORMULA.

101

IMPLICIT – not EXPLICIT. INDEPENDENT VARIABLE – ARGUMENT, INPUT, e.g. given y = f(x), x is an INDEPENDENT VARIABLE. INPUT – (value of) ARGUMENT, (value of) INDEPENDENT VARIABLE, e.g. given y = f(x),

x is INPUT; also given y = f (2), 2 is INPUT. INTEGER PART – when dividing a positive integer m into a positive integer n, k is the INTEGER PART if it is the largest positive integer producing k*m ≤ n. The REMAINDER is the difference n - k*m, e.g. when dividing 9 into 2, the INTEGER PART is 4 and the

REMAINDER is 1, so that 2

14

2

14

2

9=+= .

INVERSE (to an) OPERATION – operation that undoes what the original OPERATION does. LAST OPERATION – see ORDER OF OPERATIONS. LHS – Left Hand Side of the EQUATION or FORMULA, to the left of the '='-sign LINEAR EQUATION −– an EQUATION which involves only FREE TERMS and TERMS which contain the UNKNOWN only as a FACTOR, e.g. 2x − 3 = 0 is a LINEAR EQUATION, 2 is a COEFFICIENT in front of the UNKNOWN and −3 is a FREE TERM. LOCUS – a set of points with a specific property. NECESSARY - A is a NECESSARY condition of B if B ⇒ A (B implies A), so that B cannot take place unless A is satisfied. NON-DIMENSIONAL QUANTITY - a quantity taking any value from an allowed set of numbers. NON-LINEAR EQUATION – an EQUATION which is not LINEAR, e.g. 2 ln (x) – 3 = 0 is a NON-LINEAR EQUATION N in x. OPERATION (mathematical) – action on CONSTANTS and VARIABLES. When all CONSTANTS and VARIABLES entering an EXPRESSION are given values, OPERATIONS are used to EVALUATE this EXPRESSION. ORDER OF OPERATIONS When EVALUATING a (mathematical) EXPRESSION it is important to know the order in which to perform the OPERATIONS. By convention, the ORDER OF OPERATIONS is as follows: First, expression in BRACKETS must be EVALUATED. If there are several sets of brackets, e.g. {[( )]}, expressions inside the inner brackets must be EVALUATED first. The rule applies not only to brackets explicitly present, but also to brackets, which are implied. Two special cases to watch for are fractions and functions. Indeed, when (a + b)/(c + d) is presented as a two-storey fraction the brackets are absent, and some authors do not bracket ARGUMENTS of elementary FUNCTIONS, such as exp, log, sin, cos, tan etc. In other words, ex should be understood as exp (x), sin x as sin (x) etc. Other OPERATIONS must be performed in order of decreasing complexity, which is

102

FUNCTIONS f ( )

POWERS (including inverse operations of roots and logs)

MULTIPLICATION (including inverse operation of division)

ADDITION (including inverse operation of subtraction)

That is, the more complicated OPERATIONS take precedence. OUTPUT – (value of) DEPENDENT VARIABLE, (value of the) FUNCTION, e.g. given y = f(x),

y is OUTPUT; also given f(x) = 2x +3 and x = 2, 7 is OUTPUT (indeed, 2*2+3 = 7). PRODUCT – a (mathematical) EXPRESSION in which the LAST operation (see the ORDER OF OPERATIONS) is multiplication, e.g. ab is a PRODUCT, and so is (a + b)c. QUOTIENT - a mathematical expression where the last operation is division. RANGE – the set of all possible values of a FUNCTION. REARRANGE EQUATION, FORMULA, IDENTITY – the same as TRANSPOSE. REMAINDER – see INTEGER PART. RHS – Right Hand Side of the EQUATION or FORMULA, to the right of the = sign. ROOT OF THE EQUATION – SOLUTION of the EQUATION. SEQUENCE – a function with an INTEGER ARGUMENT. SIMPLE EQUATION – an EQUATION that can be rearranged to contain the unknown in one term only. SIMPLE TRANSFORMATION - translation, scaling or reflection. SOLUTION OF AN ALGEBRAIC EQUATION – constant values of the UNKNOWN VARIABLE which turn the EQUATION into a true statement. SOLVE – find SOLUTION of the EQUATION. SUBJECT OF THE EQUATION – the unknown is the SUBJECT OF THE EQUATION if it stands alone on one side of the EQUATION, usually, LHS. SUBSTITUTE – put in place of. SUFFICIENT - A is a SUFFICIENT condition of B if A ⇒ B (A implies B), so that if A is satisfied, then B takes place. SUM – a (mathematical) EXPRESSION in which the LAST operation (see ORDER OF OPERATIONS) is addition, e.g. a + b is a SUM, and so is a(b + c) + ed.

103

TERM – a (mathematical) EXPRESSION that is added to another (mathematical) EXPRESSION, e.g. a + b is a SUM of two TERMS, a and b. TRANSPOSE EQUATION, FORMULA, IDENTITY – make a particular unknown the subject of EQUATION, FORMULA, IDENTITY, so that it stands on its own in the LHS or RHS of the corresponding mathematical statement. UNKNOWN – a VARIABLE whose value or expression can be found by solving an EQUATION, e.g. in equation x + 2 = 3, x is an UNKNOWN. VALUE – a number. VARIABLE – a mathematical quantity that can take a range of (numerical) values and is represented by a mathematical symbol, usually a Latin letter, usually from the second part of the alphabet. If there is only one VARIABLE, the preferred choice for its symbol is x. The second choice is y and the third, z. If there are more variables, then one chooses, in the order of preference, letters u, v, w, s, t, r, p and q, then the upper case letters in the same order of preference. If more VARIABLES are required, we make use of subscripts, superscripts and Greek letters.

104

VI. STUDY SKILLS FOR MATHS

Assuming that you have an average background in mathematics you need to study these notes on your own for 6 hours each week:

1. Spend half an hour revising the Summary or Summaries suggested for Self Study. You should be able to use Order of Operations, algebraic operations and Decision Trees very fast. Do not forget to keep consulting the Glossary. 2. Spend 2.5 hours revising previous Lectures and Solutions to Exercises. 3. Spend 1.5 hours studying the latest Lecture (see tips below on how to do that). 4. Spend 1.5 hours doing the exercises given in that lecture for self-study.

Some of you need to study 12 hours a week. Then multiply each of the above figures by two!

This is how to study each new lecture:

1. Write down the topic studied and list all the subtopics covered in the lecture. Create a flow chart of the lecture. This can be easily done by watching the numbers of the subtopics covered.

For example in Lecture 1 the main topic is ALGEBRA but subtopics on the next level of abstraction are 1.1, 1.2, …, 1.3.

Thus, you can construct the flow chart which looks like that

2. Write out the glossary for this lecture; all the new words you are to study are presented in bold red letters.

3. Read the notes on the first subtopic several times trying to understand how each problem is solved

ALGEBRA

variables operations on variables general remarks glossary

whole addition, natural subtraction integer multiplication rational division

105

4. Copy the first problem, put the notes aside and try to reproduce the solution. Be sure in your mind that you understand what steps you are doing. Try this a few times before checking with the notes which step is a problem.

5. Repeat the process for each problem.

6. Repeat the process for each subtopic.

7. Do exercises suggested for self-study

Here are a few tips on how to revise for a mathematics test or exam: 1. Please study the Summaries first. 2. Keep consulting the Glossary. 3. Then study Lectures and Solutions to relevant Exercises one by one in a manner suggested above. 4. Then go over Summaries again. Here is your check list: you should be thoroughly familiar with 1. The Order of Operations Summary and how to “make invisible brackets visible”. 2. The Algebra Summary, in particular, how to remove brackets, factorise and add fractions. You should know that division by zero is not defined. Rules for logs are secondary. You should know what are integers and real numbers. You should not "invent rules" with wrong cancellations in fractions. You should not invent rules on changing order of operations, such as addition and power or function. You should all know the precise meaning of the words factor, term, sum and product. 3. The concept of inverse operation. 4. The Decision Tree for Solving Simple Equations. 5. The formula for the roots of the quadratic equation and how to use them to factorise any quadratic. 6. The diagrammatic representation of the function (see the Functions Summary). You should know that a function is an operation (or a chain of operations) plus domain. You should know the meaning of the words argument and domain. You should know what is meant by a real function of real variable (real argument). You should know the precise meaning of the word constant - you should always say "constant with respect to (the independent variable) x or t or whatever..." 7. How to do function composition and decomposition using Order of Operations 8. How to use graphs

106

9. How to sketch elementary functions: the straight line, parabola, exponent, log, sin and cos 10. The Trigonometry Summary

11. The approximate values of e ( ≈ 2.71) and π ( ≈ 3.14)

12. What is j (= 1− ) and what is j2 (= − 1) 13. The Cartesian and exponential form of a complex number and how to represent a complex number on the Argand diagram (the Complex Numbers Summary) 14. How to add, multiply, divide complex numbers, raise them to integer and fractional power 15. The Differentiation Summary 16. The Integration Summary 17. The Sketching by Simple Transformation Summary 18. The Limits Decision Tree and basic indeterminate forms 19. Sketching by analysis 20. The definition of a mean of a function on an interval

107

VII. TEACHING METHODOLOGY (FAQs) Here I reproduce a somewhat edited correspondence with one of my students who had a score of about 50 in his Phase Test and 85 in his exam. You might find it instructive. Dear Student The difference between stumbling blocks and stepping stones is how you use them! Your letter is most welcome and helpful. It is extremely important for students to understand the rationale behind every teacher's decision. All your questions aim at the very heart of what constitutes good teaching approach. For this reason I will answer every one of your points in turn in the form of a Question - Answer session: Q: I have understood the gist of most lectures so far. However there have been a number of lectures that towards the end have been more complicated and more complex methods were introduced. When faced with the homework on these lectures I have really struggled. The only way I have survived have been to look at the lecture notes, Croft’s book, Stroud's book and also various websites. A: Any new topic has to be taught this way: simple basic facts are put across first and then you build on them. If you understand simple facts then the more sophisticated methods that use them seem easy. If they do not this means that you have not reached understanding of basics. While in general, reading books is extremely important, at this stage I would advise you to look at other books only briefly and only as a last resort, spending most of the time going over the lectures over and over again. The problem with the books available at this level is that they do not provide too many explanations. LEARNING IS A CHALLENGIG AND UNINTUITIVE POCESS. IF YOU BELIEVE THAT YOU UNDERSTAND IT DOES NOT MEAN THAT YOU DO! Q: Many of the homework questions are way beyond the complexity of any examples given in lectures. Some are or seem beyond the examples given in books. A: None of them are, although some could be solved only by very confident students who are already functioning on the level of the 1st class degree. There are four important points to be aware of here: 1. If you have not reached the 1st class level yet, it does not mean that you cannot reach it in future. 2. 1st class degree is desirable to be accepted for a PhD at elite Universities, others as well as employers are quite happy with 2.1. 3. It is absolutely necessary for students to stretch themselves when they study and attempt more challenging problems than they would at exams, partly because then exams look easy. 4. Even if you cannot do an exercise yourself, you can learn a lot by just trying and then reading a solution. Q: This has been and continues to be demoralising.

108

A: A proper educational process is a painful one (no pain no gain!), but it also should be enlightening. One of the things you should learn is how to "talk to yourself" in order to reassure yourself. One of the things that I have been taught as a student and find continually helpful is the following thought: "Always look for contradictions. If you find a contradiction (that is, see that there is something fundamentally flawed in your understanding) - rejoice! Once the contradiction is resolved you jump one level up in your mastery of the subject (problem)." In other words, you should never be upset about not understanding something and teach yourself to see joy in reaching new heights. Q: Take this week as an example. Exercises suggested in Lecture 19 look to me like 'integration by parts ' which I learned in Croft's book. However in your notes these seem to be 'separate terms’. There were not any examples given in the lecture on ' integration by parts' or by 'separate terms ' and I am not sure if we were meant to be using the 'substitution method', if this is even possible. A: This example illustrates very well why at this stage reading books brings more harm than good. There are no questions in Lecture 19 which require "integration by parts", only the techniques that have been discussed in that Lecture, that is, "separate terms" rule, "constant factor out" rule and "substitution" method. It is true that there were no examples in the lectures on how to "separate terms". This omission is due to the fact that sometimes when during a lecture, I omit very simple steps. They are included into the written lectures. . Q: I also feel that the pace has been too high; we spent only about an hour on sequences and 2 weeks on integration. A: We have only 22 weeks in one year to cover the necessary material, and it is spread out accordingly. However, you do not need to know more about sequences than was already discussed. Q: When revising each week it is most unnatural to have to take your mindset back a week or two to try to remember what you have learned at a certain stage. I really do not think that many do it. A: This question touches on one of the most fundamental aims of education - development of long term memory. Both short-term memory and long-term memory are required to be a successful student and a successful professional. When I ask you to memorise something (and say that this is best done by going over the set piece just before going to bed) I am exercising your short-term memory. How can you develop a long-term memory, so that what we study to-day stays with you - in its essence - for ever? The only way to do that is by establishing the appropriate connections between neural paths in your brain. If you have to memorise a sequence of names, facts or dates there are well established techniques promoted in various books on memory. They suggest that you imagine a Christmas tree or a drive-in to your house, imagine various objects on this tree or along the drive-way and associate the names, facts or dates with these objects. However, this technique will not work with technical information. What you need to establish are much deeper - meaningful - connections. The only way to do this is to go over the same material again and again, always looking at it from a new vantage point. While your first intuitive reaction is that "it is most unnatural to have to take your mindset back a week to try to remember what you have learned at a certain stage", this is the only proper way to learn a technical subject and develop your long term memory.

109

Q: Related to this is the strange system where only the specified method can be used to derive an answer. At our level I feel that any method which produces the correct answer should be accepted. If you have been used to doing something one way and are forced to change then, for students that are a bit weak anyway, this will be a problem. A: This is actually a classical educational technique, aiming at two things at once: 1. practicing certain methods and techniques, 2. developing students' ability to "work to specs". People who do not come to terms with this idea are going to have problems with the exam questions where the desired techniques are specified. They will lose most of the marks if they use another technique. Q: The principles based on the first principles I feel are out of the question at our level in S1 and should be left to S2 as a minimum. A: These proofs are above the A-level, but are definitely the 1st year level. Everyone should be able to start these proofs on the right note, namely write out the appropriate definitions and then SUBSTITUTE the appropriate functions. Only the students who reached the 1st class level are expected to finish these proofs using various tricks, but everyone should be able to follow the proofs as given by their teachers in Solutions to Exercises and learn how the tricks work. Also, remember, that in real life you will hardly ever differentiate or integrate using differentiation or integration techniques, but remembering the definitions (a derivative is a local slope or a local rate of change and integral is a signed area between a curve and a horizontal axis) might come very handy in your engineering life. Q: Techniques should be introduced into the whole system to help build confidence, although I realise that there is a balance to be struck. A: Techniques are introduced according to the internal logic of the material. But confidence building is important and this is something teachers and students have to work at together. Teachers unfortunately have little time for that, all we can do is keep saying "good, good" when progress is made. You spend more time with yourself, so please keep reminding yourself that Exercises are only there to help to learn. What is important is that you are constantly stretching yourself. Please keep reminding yourself how much you achieved already. Surely, there are lots of things you can do now that you could not even dream of doing before. An extremely important educational point that you are touching upon here is the following: the so-called liberal system of education that was introduced in the 60s (and consequences of which we all suffer now) provided only "instant gratification". What the real education should be aiming at is "delayed gratification". You will see the benefits of what you are learning now - in their full glory - LATER, in year 2 and 3, not to-day. Hope this helps!

Basic Calculus Unveiled1 - Sound ... - Sound · PDF file2 Lecture 18. INTEGRAL CALCULUS:...

Documents

Transcript of Basic Calculus Unveiled1 - Sound ... - Sound · PDF file2 Lecture 18. INTEGRAL CALCULUS:...